Let $S=\{1,2,\cdots,2n\}$, and $S\_i \subseteq S(i=1,2,\cdots,n+1)$ be $n+1$ subsets, each of which contains half of the $2n$ elements, namely $|S\_i|=n$. Consider the following expression:
$$M=\max\_{1\le i<j \le n+1} |S\_i \cap S\_j|$$


1. When $n$ is even, it seems that $M\ge n/2$ for any choice of $S\_i$. I think it is a really interesting phenomenon, but how does one give a proof of this.
2. Also by experiment, I find that$\{(i,j):1\le i<j \le n+1,|S\_i \cap S\_j|\ge n/2\}$ is $\Omega(n)$, not only one pair.
3. This question can be generalized to the following one: Let $S=\{1,2,\cdots,n\}$, and $S\_i \subseteq S$ be $m$ subsets of $S$, each of which contains exactly $k$ elements. Can the maximal size of $|S\_i \cap S\_j|(1\le i<j \le m)$ be bounded from below?
4. This question can be changed to a real version. Consider $x\_i \in [0,1]^n$ be $m$ vectors, each of which has a fixed norm: $\| {x\_i}\|\_2=k$. Can the maximal inner product of $x\_i \cdot x\_j(1\le i<j \le m)$ be bounded from below?


Maybe the third and fourth question are hard, but if they can be solved, I think the results will be very useful. Also, the correctness of the first two questions have been checked for a few instances.



The maximal size of intersection of two sets
Here is an answer to the first point and shows how you can also do the third point.


Let $\displaystyle \bigcup\_{i=1}^{n+1} S\_i=\{b\_{1},b\_{2},\ldots, b\_{m}\}$.
We consider the following table with rows indexed by the sets $S\_1,S\_2,\ldots, S\_{n+1}$ and columns indexed by the elements $b\_1,b\_2,\ldots, b\_m$. We put a $1$ in the cell $(i,j)$ with $1\leq i\leq k$ and $1\leq j\leq m$ if $b\_j$ is an element of the set $S\_i$. Let $x\_i$ the number of $1$'s in the column $i$, or equivalently it is the number of sets in which the element $b\_i$ appears with $1\leq i\leq m$.


Now we have that $S=\displaystyle \sum\_{i=1}^{m} x\_i$ is equal to the sum of the numbers in the whole table, and doing the sum by rows we obtain that $\displaystyle \sum\_{i=1}^{m} x\_i=\sum\_{i=1}^{n+1} |S\_i|=n(n+1)$.


Let us look further at $\displaystyle \sum\_{1\leq i<j\leq n+1} |S\_i\cap S\_j|$. This is equal to the number of pairs of $1$'s who are in the same column. But this is equivalent in our notation to the fact $\displaystyle \sum\_{1\leq i<j\leq n+1} |S\_i\cap S\_j|=\sum\_{i=1}^{m} \dbinom{x\_i}{2}$.


Now just use the convexity of the function $\binom{x}{2}$ to get that 
$$\sum\_{i=1}^{m} \dbinom{x\_i}{2}\geq m \dbinom{\frac{n(n+1)}{m}}{2}=\cfrac{(n(n+1)-m)n(n+1)}{2m}$$.


Since there are $\dbinom{n+1}{2}$ intersection the maximum size of the intersection is at least $\cfrac{n(n+1)-m}{m}$. Finally $m\leq 2n$ and you get the desired size almost namely $\cfrac{n-1}{2}$.


