Finding the minimal number of members I've been working on the following problem

For every issue in the Blue's association, a commission with 10 members (belonging the Blue's) is formed in order to solve the problem. The only condition is

There can't be two commissions having more than one member in common

The Blue's association has formed this year 40 commissions.
What's the minimal amount of members in the Blue's association?

I've only found out the following

For any commission you can form $\binom{10}{2}=45$ different pairs and none of them can appear in another commission.
Since 40 different commissions are formed, the minimal number of pairs is $45\times 40=1800$.
Denote by $n$ the number of members. Thus $$\binom{n}{2}≥1800\Rightarrow n>60$$

$$$$

The minimal amount of members has to be 100 or less.

You can observe a distribution for 100 members here

My question:
Is 100 the answer or is there an ever smaller possible amount of members?
If so, how can I prove it?
 A: This is just a partial answer. I will show that $85$ members suffice; I don't know if $85$ is the minimum.
Recall that a projective plane of order $n$ exists if $n$ is a prime power: it has $n^2+n+1$ points and $n^2+n+1$ lines; each line has $n+1$ points, and there are $n+1$ lines through each point; any pair of lines meets in a unique point, and any pair of points determines a unique line.
Consider a projective plane of order $9$; it has $9^2+9+1=91$ points and $91$ lines; there are $10$ points on each line and $10$ lines through each point. A set of points is in general position if no three of the points are collinear. Note that, if we have a set of $t$ points in general position, then the lines determined by those points (taken two at a time) cover a total of at most $t+8\binom t2$ points; as long as $t\le5$ then the number of covered points is at most $5+8\binom52=85\lt91$, so we can add another point to the set and still have them in general position. Thus we can find a set $S$ of $6$ points in general position.
Let the members of the Blue's association be the $91-6=85$ points which are not in $S$. The commissions are the lines which do not meet $S$; they have $10$ members each, and any two have exactly one member in common. Finally, by the in-and-out formula, the number of commissions is
$$91-\binom61\cdot10+\binom62\cdot1=46.$$
P.S. Let $m$ be the minimum possible number of members. I showed above that $m\le85$. On the other hand, I have a small improvement on your lower bound $m\ge61$.
Suppose the $i^\text{th}$ member belongs to $d_i$ commissions; then
$$\sum_{i=1}^md_i=400$$
since there are $40$ commissions with $10$ members each. Moreover $d_i\le9$ since $m\le85\lt91$. Let $k=|\{i:d_i\ge5\}|$. Then
$$400=\sum_{i=1}^md_i\le4(m-k)+9k=4m+5k\le340+5k,$$
whence $k\ge12$; i.e., there are at least $12$ members who are on at least $5$ commissions. Choose two members $i$ and $j$ who are on at least $5$ commissions.
Case 1. There is a commission containing both $i$ and $j$.
First, there are $10$ members on the commission which $i$ and $j$ both belong to. Next $i$ belongs to $4$ more commissions, with $36$ additional members. Finally, $j$ belongs to $4$ more commissions, each of which contains at most one member of each of the $5$ commissions containg $i$, and at least $5$ members who haven't been counted yet, for a total of $20$ new members. This shows that $m\ge10+36+20=66$.
Case 2. There is no commission containing both $i$ and $j$.
In this case a similar argument shows that $m\ge67$.
This proves that $m\ge66$. Combining this with the upper bound shown earlier, we have
$$66\le m\le85.$$
A: This post exhibits a solution with $82$ members.  Combined with the excellent answer by @Song, this means $82$ is indeed optimal.
Motivation: The excellent answer by @Song and the followup comments by @Servaes make me wonder... perhaps if we look for 41 commissions (not 40) then there is a solution with a great deal of symmetry:


*

*(a) 82 members (the optimal answer we seek)

*(b) 41 commissions (exceeds OP requirement)

*(c) each member associated with exactly 5 commissions (not part of OP)

*(d) each commission associated with exactly 10 members (equals OP requirement)

*(e) each 2 members have exactly 1 common commission (not part of OP)

*(f) each 2 commissions have exactly 1 common member (exceeds OP requirement)


This would be like a finite projective plane, but with 82 points and 41 lines.  However, in a finite projective (respectively: affine) plane, the no. of points and no. of lines are equal (respectively: almost equal), and this is probably why a solution based on FPP only reaches 84.  So I decided to look at related structures called Block Designs, Steiner Systems, etc. where there are typically many more "lines" than "points".  After quite a bit of digging, I think I found the right structure:
The solution:  It is a Steiner $S(t=2,k=5,n=41)$ system.  A Steiner system is defined by the following properties:


*

*there are $n=41$ objects (these are the commissions)

*there are $b$ blocks (these are the members), each block (member) being a subset of objects (i.e. the commissions he/she is associated with)

*each block has $k=5$ objects (each member is associated with 5 commissions)

*every $t=2$ objects is contained in exactly 1 block (every 2 commissions have exactly 1 common member)
So this already satisfies (b), (c) and (f).  Next, quoting from https://en.wikipedia.org/wiki/Steiner_system#Properties we have:


*

*$b = {n \choose t} / {k \choose t} = (41 \times 40) / (5 \times 4) = 41 \times 2 = 82$, satisfying (a)

*$r = {n-1 \choose t-1} / {k-1 \choose t-1} = 40 / 4 = 10$, where $r$ denotes "the number of blocks containing any given object", i.e. the number of members associated with any given commission, satisfying (d).
Thinking more, I don't think (e) can be satisfied.  However, (e) is not needed for the OP, so it doesn't matter.
So finally we just need to prove that such a Steiner $S(t=2,k=5,n=41)$ system exists.  This existence is non-trivial, but luckily more digging reveals: 


*

*https://math.ccrwest.org/cover/steiner.html has a list of Steiner systems that are known to exist.  $S(2,5,41)$ (the webpage sometimes lists the 3 parameters in different order) is not part of any infinite families listed, but if you go further down the page it is listed as a standalone example; clicking on that link goes to...

*https://math.ccrwest.org/cover/show_cover.php?v=41&k=5&t=2 which exhibits the system, created via "Cyclic construction" whatever that means.
I did not check the numbers thoroughly, but if I understand the webpage correctly, there should be 82 rows (members / blocks), each containing 5 numbers (commissions), all the numbers being 1 through 41 inclusive (the 41 commissions), each number (commission) should appear in 10 rows, and every pair-of-numbers should appear in 1 row.
I am not an expert in any of these, so if I had a mistake or misunderstanding above, my apologies.  Perhaps someone more expert can check my work?
A: Here's a partial answer that increases the lower bound for any (not necessarily optimal) solution to $m\ge 74$.
Suppose there is a solution with $m$ members and we know that there are two members each in $l+1$ commissions, then
$$m\ge 9(l+1)+(8-l)(l+1)+2.$$
This is because if member one is in $\ge l+1$ commissions, each commission needs to be filled with $9$ new members since these $l+1$ commissions already have maximum overlap. For the commissions that member two is in, each needs $9$ more members to be accounted for. We can't have any overlap between these commissions because they already have maximum overlap (that being member two). We can at best choose one member from each group with member one in it, giving us $l+1$ members, but the other $9-(l+1)=8-l$ are new. This gives $9(l+1)+(8-l)(l+1)$ members other than the two we started with. (Note that this is the best bound in $l$ possible).
Now, suppose $m$ members has a solution to the problem. Let $d_i$ be the how many commissions the $i$-th member is in. First note that $m\ge 9d_i+1$ for every $i$, so $d_i\le \lfloor (m-1)/9\rfloor$. Let $k_l=|\{i\; :\; d_i>l\}|$. Then
$$400=\sum_{i=1}^md_i\le l(m-k_l)+\lfloor (m-1)/9\rfloor k_l.$$
Hence
$$k_l\ge \frac{400-lm}{\lfloor (m-1)/9\rfloor -l}.$$
Since $k_l$ is an integer, if $\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$ then $k_l\ge 2$, meaning that there is at least $2$ members in at least $l+1$ commissions so by the above $m\ge 9(l+1)+(8-l)(l+1)+2$. Note that $\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$ exactly when $\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$. Hence we have that for all
$$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$
we have
$$m\ge 9(l+1)+(8-l)(l+1)+2.$$
For this to be satisfied gives
$$m\ge 74.$$
A: Let $i$ denote each member of Blue's association and assume that there are $N$ members in total, that is, $i=1,2,\cdots, N.$ And let $j,k=1,2,\ldots, 40$ denote each of 40 commission. We will show that $N$ is at least $82$. 
Consider the set
$$
S=\{(i,j,k)\;|\;1\leq i\leq N, 1\leq j<k\leq 40, i\text{ belongs to }j,k\text{-th commission.}\}.
$$ Let $d_i$ denote the number of commissions that $i$ joined. We will calculate $|S|$ using double counting method. First, note that
$$
|S|=\sum_{(i,j,k)\in S}1 = \sum_{1\leq j<k\leq 40} \sum_{i:(i,j,k)\in S}1\leq \sum_{1\leq j<k\leq 40}1=\binom{40}{2},
$$ since for each $j<k$, there is at most one $i$ in common. On the other hand,
$$
|S| = \sum_{1\leq i\leq N} \sum_{(j,k):(i,j,k)\in S}1 = \sum_{1\leq i\leq N} \binom{d_i}{2},
$$ since for each $i$, the number of pairs $(j,k)$ that $i$ joined is $\binom{d_i}{2}$.
We also have $$\sum_{1\leq i\leq N}d_i = 400,$$by the assumption.
Finally, note that the function $f(x)= \binom{x}{2} = \frac{x^2-x}{2}$ is convex. Thus by Jensen's inequality we have that
$$
\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq Nf\left(\frac{\sum_i d_i}{N}\right)=N\binom{\frac{400}{N}}{2}.
$$ This gives us the bound
$$
40\cdot 39 \geq 400\cdot(\frac{400}{N}-1),
$$and hence
$$
N \geq \frac{4000}{49} = 81.63\cdots
$$ This establishes $N\geq 82$. However, I'm not sure if this bound is tight. I hope this will help.
$\textbf{Note:}$ If $N=82$ is tight, then above argument implies that $d_i$'s distribution is almost concentrated at $\overline{d} = 400/82 \sim 5$. 
EDIT: @antkam's answer seemingly shows that $N=82$ is in fact optimal.
