6.17 Suppose there are $n$ tennis players who are to play doubles matches with the rule that each player must play exactly one match against every other player as an opponent. For which natural numbers $n$ can such a tournament be organized?
[Solution] $n$ satisfies $n \equiv 1(\bmod 8)$.
Assume the tournament has been conducted as required. Thus, for tennis player $A$, his opponents must be paired. Therefore, $n$ must be an odd number. Additionally, according to the problem, $A$ participates in $\frac{1}{2}(n-1)$ matches. Therefore, $n$ people participate in $\frac{1}{2} n(n-1)$ matches (including repeated counts). Since each match involves 4 people, in the above counting process, each match is counted exactly 4 times. Thus, $\frac{1}{2} n(n-1)$ should be a multiple of 4, i.e., $n-1$ is a multiple of 8.

Next, we prove that for $k=1,2, \cdots$, the doubles tournament for $n=8k+1$ players can be conducted as required.

When $k=1$, we place 9 players at the vertices of a regular nonagon $A_{1} A_{2} \cdots A_{9}$.

First, select $A_{1} A_{2}$ and $A_{3} A_{5}$ to play a doubles match, and connect the opponents with line segments. Then, we rotate these line segments by multiples of $\frac{2 \pi}{9}$ to get the structure of the other eight matches. In this structure, each chord $A_{ij}$ appears exactly once.

For $k>1$, we first select one player from the $8k+1$ tennis players, then divide the rest into $k$ groups of 8 people each, and sequentially conduct the matches for each group and the selected player according to the above procedure for 9 people. The remaining matches are between opponents from different groups, which can be achieved by dividing each group into four teams (each team of two people) and then conducting matches between different groups.