Tokarev.i.

In a tennis tournament, $n$ participants want to play doubles (two against two) matches so that each participant has every other participant as an opponent exactly once. For which $n$ is such a tournament possible?
Let the tournament described in the problem be held. Then all opponents of one tennis player are divided into pairs, so $n$ is odd. All possible pairs of opponents are divided into quadruples of pairs that played in the same match. Therefore, the number $1 / 2 n(n-1)$ of these pairs is divisible by 4, from which $n-1=8 k$.

We will prove that for any natural $k$, the specified tournament for $n=8 k+1$ participants is possible. For $k=1$, to describe the tournament, we will correspond the tennis players to the vertices of a regular nonagon $A_{1} A_{2} \ldots A_{9}$. In the figure, the match between the pair $\left(A_{1}, A_{2}\right)$ and the pair $\left(A_{3}, A_{5}\right)$ is depicted, with segments connecting the opponents.

![](https://cdn.mathpix.com/cropped/2024_05_06_7e3e695203bdf41f05f0g-24.jpg?height=303&width=323&top_left_y=1452&top_left_x=868)

By rotating this construction of segments around the center by angles that are multiples of $2 \pi / 9$, we obtain images for the remaining eight matches. In this process, each chord of the form $A_{i} A_{j}$ is used exactly once, since it is exactly one of the chords $A_{2} A_{3}, A_{1} A_{3}, A_{2} A_{5}$, and $A_{1} A_{5}$.

For $k>1$, we select one of the $8 k+1$ tennis players and divide the rest into $k$ groups of 8 people each.

By adding the selected tennis player sequentially to each group, we will conduct tournaments in them according to the described scheme for 9 people. Then it remains only to conduct matches between opponents from different groups. For this, it is sufficient to divide each group into four teams of two people and conduct all possible matches between teams from different groups.

## Answer

$n=8 k+1$, where $k \in \mathbf{N}$.