Belarusian Mathematical Olympiad

Number girls and boys by numbers from $1$ to $2n$. For each $i$-th girl we correspond a vector $S_i$ with $j$-th entry equal to $1$ if the $i$-th girl has a dance with the $j$-th boy, and equal to $-1$ otherwise. Then the condition is equivalent to $S_i \cdot S_j^T = 0$ for every $i \neq j$, because $S_i$ and $S_j$ differ at exactly $n$ positions.

Define the $2n \times 2n$ matrix $S$ with rows $S_i$, $i = 1...2n$. Then $S \cdot S^T = 2nI$, where $I$ is the identity $2n \times 2n$ matrix. This implies, that $\frac{1}{2n}S^T = S^{-1}$. Therefore $S^T \cdot S = 2nI$, which means that every two columns differ at exactly $n$ positions, which is equivalent to the required statement.