In a tennis tournament, $n$ players want to make $2$ vs $2$ matches such that each player has each of the other players as opponents exactly once. Find all possible values of $n$.
To solve the problem, we need to determine the values of \( n \) such that \( n \) players can be arranged into \( 2 \) vs \( 2 \) matches, with each player playing against every other player exactly once. 

1. **Counting the Number of Games:**
   - Each player plays against every other player exactly once. Therefore, each player will participate in \(\frac{n-1}{2}\) games (since each game involves 4 players, and each player will be paired with every other player exactly once).
   - The total number of games can be calculated by considering that each game involves 4 players. Thus, the total number of games is \(\frac{n(n-1)}{8}\).
   - For this to be an integer, \(\frac{n(n-1)}{8}\) must be an integer. This implies that \(n(n-1)\) must be divisible by 8.

2. **Divisibility Condition:**
   - Since \(n\) and \(n-1\) are consecutive integers, one of them must be even. For \(n(n-1)\) to be divisible by 8, we need to ensure that \(n\) or \(n-1\) is divisible by 8.
   - This implies that \(n \equiv 1 \pmod{8}\).

3. **Construction:**
   - Let \(n = 8k + 1\) for some integer \(k\).
   - Number the players as \(\{1, 2, \ldots, 8k+1\}\).
   - For each \(\ell = 1, 2, \ldots, 8k+1\), consider the following 4-tuples where elements are taken such that \(n+1 \equiv 1\):
     \[
     (\ell, \ell+4k+1 \, ; \, \ell+1, \ell+2), \quad (\ell, \ell+4k+1 \, ; \, \ell+3, \ell+4), \quad \ldots, \quad (\ell, \ell+4k+1 \, ; \, \ell+2k-1, \ell+2k)
     \]
   - This construction ensures that each player plays against every other player exactly once.

4. **Verification:**
   - There are exactly \(k(8k+1)\) 4-tuples/matches, and they are symmetric in each player.
   - To verify, check that player 1 plays against every opponent exactly once. This is clear from the construction.

5. **Example:**
   - For \(n = 9\), the construction gives:
     \[
     (1, 6, 2, 3), \quad (2, 7, 3, 4), \quad (3, 8, 4, 5), \quad (4, 9, 5, 6), \quad (5, 1, 6, 7), \quad (6, 2, 7, 8), \quad (7, 3, 8, 9), \quad (8, 4, 9, 1), \quad (9, 5, 1, 2)
     \]

The final answer is \( \boxed{ n = 8k + 1 } \) for all \( k \in \mathbb{N} \).