SOLVED - $500

Suppose that we have a family $\mathcal{F}$ of subsets of $[4n]$ such that $\lvert A\rvert=2n$ for all $A\in\mathcal{F}$ and for every $A,B\in \mathcal{F}$ we have $\lvert A\cap B\rvert \geq 2$. Then
\[\lvert \mathcal{F}\rvert \leq \frac{1}{2}\left(\binom{4n}{2n}-\binom{2n}{n}^2\right).\]

Conjectured by Erdős, Ko, and Rado [ErKoRa61]. This inequality would be best possible, as shown by taking $\mathcal{F}$ to be the collection of all subsets of $[4n]$ of size $2n$ containing at least $n+1$ elements from $[2n]$.

Proved by Ahlswede and Khachatrian [AhKh97], who more generally showed the following. Let $2\leq t\leq k\leq m$ and let $r\geq 0$ be such that \[\frac{1}{r+1}\leq \frac{m-2k+2t-2}{(t-1)(k-t+1)}< \frac{1}{r}.\] The largest possible family of subsets of $[m]$ of size $k$, such that the pairwise intersections have size at least $t$, is the family of all subsets of $[m]$ of size $k$ which contain at least $t+r$ elements from $\{1,\ldots,t+2r\}$.