Minimum possible value of n? 
There are $100$ countries participating in an olympiad. Suppose $n$ is a
  positive integer such that each of the $100$ countries is willing to
  communicate in exactly $n$ languages. If each set of $20$ countries can
  communicate in at least one common language, and no language is common
  to all $100$ countries, what is the minimum possible value of $n$?

I tried thinking of various cases where we can get minimum value, but I am not sure and they are very different. 
My opinion- let 99 countries have a common language, so all the groups of 20 countries including the left out country will have to speak different language,so that the rule follows. So the n= (99C19) +1
 A: *

*We have $100$ sets of languages $S_i$ of size $|S_i|\leq n$ each and for atleast one $i$ we have $|S_i|=n$.

*We know for each draw $d$ of $20$ countries, we have at least one common language:
$$
\left\lvert \bigcap_{j=1}^{20} S_{d_j} \right\rvert \ge 1
$$

*And there is no language common to all countries:
$$
\bigcap_{i=1}^{100} S_i = \emptyset
$$
This is slightly different than the question, but if we got sets with these properties we can just fill them up with random disjoint languages to get $|S_i|=n$ for all $i$, without removing any of the 2 other properties. So if we find an $n$ and sets $S_i$ with these properties this $n$ also fulfills the properties of the question.
Let $n$ be the minimal number s.t. we can find sets $S_i$ with the properties 1.,2.,3. from above, and let $S_i$ (for $i=1,...,100$) be such sets.
I claim if $m=\left\lvert \bigcup_{j=1}^{100} S_{j} \right\rvert$, the number of differet languages in total, is greater than $n+1$, we can find sets $A_i$ with the properties 1.,2.,3. and $\left\lvert \bigcup_{j=1}^{100} A_{j} \right\rvert=m-1$. Property 1. $\rightarrow\exists i$ s.t. $|S_i|=n$. W.l.o.g. $i=1$ and $\bigcup_{j=1}^{100} S_{j}=\{1,2,3,...,m\}$ and $S_1=\{1,2,3,...,n\}$. Now let for all $i=1,2,...,100$: $A_i=S_i$ if $m\notin S_i$ and $A_i=(S_i\cup\{n+1\})-\{m\}$ else. Now 1.,2.,3. is still clearly fulfilled by the sets $A_i$, so my claim follows.
So we can iterate this till $m=n+1$. Now again let $n$ be minimal again and assume $n<20$ and let $S_i$ be sets s.t. 1.,2.,3. holds and $n+1=\left\lvert \bigcup_{j=1}^{100} S_{j} \right\rvert$ (exists because of the established claim). Again w.l.o.g. $\bigcup_{j=1}^{100} S_{j}=\{1,2,3,...,n+1\}$. But now because of 3. we can find $d_1,...d_{n+1}$ s.t. $1\notin S_{d_1},2\notin S_{d_2},3\notin S_{d_3},...$ but then $$
\left\lvert \bigcap_{j=1}^{n+1} S_{d_j} \right\rvert = 0
$$
a contradiction because $n+1\leq 20$.
So $n\geq 20$. 
For $n=20$ we can define the sets $S_i=\{1,2,...,21\}-\{i\}$ for $i=1,...,21$ and $S_i=S_1$ for $i=22,...,100$, which fullfills all properties. So $20\geq n$.
So $n=20$
Idea:
If we got a solution with $m$ languages and each country speaks less or equal to $n$ languages and atleast one speaks exaclty $n$ languages and $m>n+1$ we can construct a solution with $m-1$ languages.
Then I show that a solution with $n<20$ and $m=n+1$ can't exist.
Then I showed a construction for a solution with $n=20$ and $m=n+1$.
A: Each country can speak $20$ languages.  You need $21$ languages.  A construction to show it is possible is to number the countries from $0$ to $99$, number the languages from $0$ to $20$ and have each country speak all the languages except the one that is their country number $\bmod 21$.  Any collection of $20$ countries can only prohibit $20$ of the languages being used, so there is at least one that can be used.  To show $n \lt 20$ is not possible, take the first country as the first of the group.  Add to the group a country that does not speak each language spoken by the first country.  This will be a group of at most $20$ that cannot find a common language.
A: It just asks minimum, why is there a need for combination?
Just rewrite the text as inequality with respect to n.
$\text{20 countries} = \text{1 language}\\
\text{1 country} = \dfrac{1}{20} \text{language} \\ 
\text{100 country} = \dfrac{100}{20} = \text{5 language} \\ 
\implies n \ge 5 \\ 
\implies \mathrm{min}(n) = 5
$
A: Trying to model this task:


*

*We have $100$ sets of languages $S_i$ of size $n$ each.

*We know for each draw $d$ of $20$ countries, we have at least one common language:
$$
\left\lvert \bigcap_{j=1}^{20} S_{d_j} \right\rvert \ge 1
$$

*And there is no language common to all countries:
$$
\bigcap_{i=1}^{100} S_i = \emptyset
$$
which implies $n > 1$.
No idea yet, what discrete maths problem might fit here.
