Vasilev N.B.

For what $n>1$ can it happen that in a company of $n+1$ girls and $n$ boys, all girls are acquainted with a different number of boys, while all boys are acquainted with the same number of girls?
A girl can be acquainted with $0,1, \ldots, n$ boys. Since there are $n+1$ girls and $n+1$ options, all these options are realized. In this case, the total number of pairs of acquaintances is $(0+1+\ldots+n)=n(n+1) / 2$, so each boy is acquainted with ${ }^{(n+1) / 2}$ girls. Therefore, $n$ is odd.

For any odd $n=2 m+1$, the required acquaintances can be achieved: we divide the girls into pairs (a total of $m+1$ pairs); the first girl of the $k$-th pair $(k=0,1, \ldots, m)$ will be acquainted with $k$ boys, and the second girl with the remaining $n-k$ boys. In this case, each boy is acquainted with exactly one girl from each pair.

## Answer

For any odd $n>1$.