determine if solution is correct for this problem , An $ (n, k) \minus{}$ tournament is a contest with $ n$ players held in $ k$ rounds such that:

$ (i)$ Each player plays in each round, and every two players meet at most once.
$ (ii)$ If player $ A$ meets player $ B$ in round $ i$, player $ C$ meets player $ D$ in round $ i$, and player $ A$ meets player $ C$ in round $ j$, then player $ B$ meets player $ D$ in round $ j$.

Determine all pairs $ (n, k)$ for which there exists an $ (n, k) \minus{}$ tournament.

[i]
, solution , To determine all pairs \((n, k)\) for which there exists an \((n, k)\)-tournament, we need to analyze the given conditions:

1. Each player plays in each round, and every two players meet at most once.
2. If player \(A\) meets player \(B\) in round \(i\), player \(C\) meets player \(D\) in round \(i\), and player \(A\) meets player \(C\) in round \(j\), then player \(B\) meets player \(D\) in round \(j\).

### Step 1: Analyze the first condition
The first condition implies that in each round, the players are paired up. Therefore, \(n\) must be even, because an odd number of players cannot be paired up without leaving one player out.

### Step 2: Analyze the second condition
The second condition is a bit more complex. It states that the structure of the pairings must be consistent across rounds. This condition implies that the tournament can be represented as a set of permutations of the players such that the structure of the pairings is preserved.

### Step 3: Consider small values of \(n\)
Let's start by considering small even values of \(n\) and see if we can construct a tournament for each.

#### Case \(n = 2\)
For \(n = 2\), there is only one round, and the two players meet each other. This satisfies both conditions. Therefore, \((2, 1)\) is a valid tournament.

#### Case \(n = 4\)
For \(n = 4\), we can have two rounds. In the first round, we can pair the players as \((1, 2)\) and \((3, 4)\). In the second round, we can pair them as \((1, 3)\) and \((2, 4)\). This satisfies both conditions. Therefore, \((4, 2)\) is a valid tournament.

#### Case \(n = 6\)
For \(n = 6\), we can have three rounds. In the first round, we can pair the players as \((1, 2)\), \((3, 4)\), and \((5, 6)\). In the second round, we can pair them as \((1, 3)\), \((2, 5)\), and \((4, 6)\). In the third round, we can pair them as \((1, 4)\), \((2, 6)\), and \((3, 5)\). This satisfies both conditions. Therefore, \((6, 3)\) is a valid tournament.

### Step 4: Generalize the pattern
From the above cases, we can see a pattern emerging. For \(n = 2m\) players, we can construct a tournament with \(m\) rounds. In each round, we can pair the players such that each player meets a different player each time, and the structure of the pairings is consistent.

### Step 5: Verify the general pattern
To verify that this pattern holds for any even \(n\), we can use the concept of a round-robin tournament. A round-robin tournament is a tournament where each player plays against every other player exactly once. For \(n\) players, a round-robin tournament can be represented as a complete graph \(K_n\) where each edge represents a game.

In a round-robin tournament with \(n\) players, the number of games is \(\binom{n}{2} = \frac{n(n-1)}{2}\). Since each round consists of \(\frac{n}{2}\) games, the number of rounds \(k\) is:
\[
k = \frac{\binom{n}{2}}{\frac{n}{2}} = \frac{n(n-1)}{2} \cdot \frac{2}{n} = n-1
\]

Therefore, for any even \(n\), there exists an \((n, n-1)\)-tournament.

### Conclusion
The pairs \((n, k)\) for which there exists an \((n, k)\)-tournament are:
\[
\boxed{(n, n-1) \text{ where } n \text{ is even}}
\]