, 2024

We shall prove that the answer is $n = 1$.

To begin with, note that the problem statement implies that the boys like $1, 2, \ldots, 10$ girls in some order, so there exists a boy that likes all the girls. Analogously, there must exist a girl that likes all the boys, so putting the two together always yields an admissible couple.

In the second part of the solution, we shall construct a configuration where it's impossible to form more than one such couple. Number the boys and the girls by numbers $1, \ldots, 10$ and suppose that the boy $i$ likes the girl $j$ if and only if $j \ge i$, while the girl $j$ likes a boy number $i$ if and only if $i = 1$ or $i > j$ (see the diagram below for the case $i = 5$, with boys on top and girls on bottom). With such an assignment, the $i$-th boy likes $11 - i$ girls and the $j$-th girl likes $11 - j$ boys. Also, it

is clear that the first boy is the only one that can be paired up with a girl that he likes so that she also likes him back, hence it is impossible to form two disjoint admissible couples and we are done.