11.7. In a tennis round-robin tournament, there are $n$ players participating in doubles matches (2 vs 2), and each player must face every other player as an opponent exactly once in the matches. Question: For what values of $n$ is such a round-robin tournament impossible?
11. 7. $\quad n=8 k+1, k \in N$.

Assume that the round-robin tournament can take place. Then all opponents of a tennis player can be divided into several "pairs", so $n$ is odd. All players can form "pairs" that can be divided into several "two pairs" that can play a match, so the number of "pairs" that $n$ people can form, $\frac{n(n-1)}{2}$, is a multiple of 4, hence $n-1=8 k$.

Next, we prove that for any $k \in N, n=8 k+1$ players can hold the round-robin tournament as required by the problem.

When $k=1$. To describe the round-robin tournament, we correspond the tennis players to the vertices of a regular nonagon $A_{1} A_{2} \cdots A_{9}$. In the figure on the right, it shows the match between $A_{1}, A_{2}$ and $A_{3}, A_{5}$, and the opponents are connected by line segments. By rotating the structure formed by the line segments around the center of the regular nonagon by $m \cdot \frac{2 \pi}{9}$ radians (where $m$ is a positive integer, $1 \leqslant m \leqslant 8$), we get the diagrams for the other 8 matches. At the same time, a chord like $A_{i} A_{j}$ appears once in the obtained diagrams, i.e., one of the chords $A_{2} A_{3}$, $A_{1} A_{3}$, $A_{2} A_{5}$, $A_{1} A_{5}$.

When $k>1$, from the $8 k+1$ players, we select 1 player and divide the rest into $k$ groups, each with 8 people. By adding the selected player to each group in turn, we hold the above-mentioned round-robin tournament for these 9 people. It only remains to play the matches between opponents from different groups. For this, we just need to divide each group into 4 teams, each with 2 people, and play all possible matches between the teams.