$n$-element subsets of the set $\{1,2, ..., 2n\}$. There are $\binom{2n} {n}$, $n$-element subsets of the set $\{1,2, ..., 2n\}$. I am studying the question whether one can choose from these subsets $\frac12\binom{2n} {n}$ subsets such that:
i) each element from $1$ to $2n$ belongs to exactly $\frac14 \cdot\binom{2n} {n}$ of the selected subsets;
ii) any two selected subsets had at least one common element.
If $n$ is a power of two, then the answer is no, because then $\binom{2n} {n}$ is not divisible by $4$. 
If $n$ is not a power of two, then $\binom{2n} {n}$ is divisible by $4$. For $n = 3$ I found an example:
$$\{1,2,3\},\{1,2,4\},\{1,3,5\},\{1,4,6\},\{1,5,6\}$$ 
$$\{2,3,6\},\{2,4,5\},\{2,5,6\},\{3,4,5\},\{3,4,6\}$$ 
I do not know what the answer is for other $n$, for example, when $n = 9$. 
Computer check is too big.
 A: Here is a method for when $2n-1$ is prime, and $2n\choose n$ is a multiple of 4.
Let $M=\{1,2,\ldots,2n-1\}$.
Let $B$ be the collection of $(n-1)$-element subsets of $M$.
$C$ will be a subset of half the sets in $B$, as follows:
Start with $C$ empty.  Pick any $b\in B$ that has not been chosen.  Put it in $C$, along with its offsets $b_i=\{g+i\pmod{2n-1}:g\in b\}$.  For example, if $2n-1=11$, and $b=\{1,2,4,6,7\}$, then you also include in $C$:
$$\{2,3,5,7,8\},\{3,4,6,8,9\},\{4,5,7,9,10\},\{5,6,8,10,11\},\{6,7,9,11,1\},\{7,8,10,1,2\},\{8,9,11,2,3\},\{9,10,1,3,4\},\{10,11,2,4,5\},\{11,1,3,5,6\}$$
So you put $2n-1$ sets at a time into $C$.  Keep adding sets from $B$ into $C$ until you have exactly half.  Each number from $1$ to $2n-1$ appears in the same number of sets in $C$.  We will later include $2n$ in each of these sets.
Let $D$ be the collection of $n$-element subsets of $M$, none of which is the complement of a $c\in C$.  Each number from $1$ to $2n-1$ also appears in the same number of elements of $D$.  
Your set is $\{c\cup\{2n\}:c\in C\}\cup D$
All the numbers from $1$ to $2n-1$ appear equally in $C$, and in $D$.  $2n$ appears in half the sets; so all appear in exactly half the sets.
If two sets come from the $C$ side, they have $2n$ in common.
If two sets come from the $D$ side, they have $2n$ elements from the numbers in $M$, so some element is in both sets.
If a set comes from $C$ and one from $D$, recall that $C$ and $D$ were chosen so that $c\cup d\neq M$, so they have an element in common.
