IMO 1988 Problem 2

Let $ n$ be a positive integer. Let $ A_1, A_2, \ldots, A_{2n+1}$ be sets having $2n$ elements each such that any two of them have exactly one element in common while every element of their union belongs to at least two of the given sets. For which $n$ can one assign to every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly $n$ zeros?

======================================

Let the elements be $1,2,\dots$, and let $j$ be in $m_j\ge2$ sets. Double count $(A_i,j)$ where $j\in A_i$ and we have $$2n(2n+1)=\sum_j m_j.$$

Double count $(A_i,A_k,j)$ where $j\in A_i\cap A_k$ and we have

$$\binom{2n+1}{2}=\sum_j\binom{m_j}{2},$$ or $$2n(2n+1)=\sum_j m_j(m_j-1).$$

Since $m_j-1\ge1$, we see that $m_j=2$ for every $j$ in the union of all sets. Equivalently, we associate each number the pair of sets that it belongs to. Let $f(j)\in\{0,1\}$ be the assignment.

Double count $(A_i, j)$ where $j\in A_i$ and $f(j)=1$ we get $$n(2n+1)=2\big|\left\{j:f(j)=1\right\}\big|.$$Hence $n$ must be even. We set

$$f(A_i\cap A_k)=\begin{cases}1&\big|(i-k\bmod 2n+1)\big|\le \frac{n}{2},\\0&\text{otherwise}.\end{cases}$$Intuitively, we arrange $A_1,\dots,A_{2n+1}$ on a circle and let $f(j)=1$ if and only if one of the arcs between the unique pair $(A_i,A_k)$ of sets that $j$ belongs to has less than $\frac{n}{2}$ other sets.