In a $k$-player tournament for $k > 1$, every player plays every other player exactly once. Find with proof the smallest value of $k$ such that it is possible that for any two players, there was a third player who beat both of them.
1. **Establishing the requirement:**
   We need to find the smallest value of \( k \) such that in a \( k \)-player tournament, for any two players, there exists a third player who beat both of them. 

2. **Analyzing the minimum losses:**
   Suppose \( k \) players are labeled as \( P_1, P_2, \ldots, P_k \). Each player plays against every other player exactly once, resulting in \( \binom{k}{2} \) games. We need to ensure that for any two players \( P_i \) and \( P_j \), there exists a player \( P_m \) who has beaten both \( P_i \) and \( P_j \).

3. **Minimum losses per player:**
   Consider a player \( P \). If \( P \) loses fewer than 3 games, then there are at least \( k-3 \) players that \( P \) has beaten. If \( P \) loses only to players \( A \) and \( B \), then \( A \) and \( B \) must have lost to some common player \( C \) who beats both \( A \) and \( B \). This implies that each player must lose at least 3 games to ensure that there is a player who beats any pair of players.

4. **Lower bound on \( k \):**
   If each player loses at least 3 games, then the total number of losses in the tournament is at least \( 3k \). Since each game results in one loss, the total number of games is \( \binom{k}{2} \). Therefore, we need:
   \[
   \binom{k}{2} \geq 3k
   \]
   Simplifying this inequality:
   \[
   \frac{k(k-1)}{2} \geq 3k
   \]
   \[
   k(k-1) \geq 6k
   \]
   \[
   k^2 - k \geq 6k
   \]
   \[
   k^2 - 7k \geq 0
   \]
   \[
   k(k-7) \geq 0
   \]
   Since \( k > 1 \), the smallest integer solution is \( k = 7 \).

5. **Constructing a valid tournament for \( k = 7 \):**
   We need to construct a tournament with 7 players where for any two players, there exists a third player who beats both. The table provided in the solution shows such a construction:

   \[
   \begin{array}{c|ccc|ccc}
   & \multicolumn{3}{c}{\text{wins against}} & \multicolumn{3}{c}{\text{loses against}} \\
   \hline
   1 & 5 & 6 & 7 & 2 & 3 & 4 \\
   2 & 1 & 4 & 7 & 3 & 5 & 6 \\
   3 & 1 & 2 & 5 & 4 & 6 & 7 \\
   4 & 1 & 3 & 6 & 2 & 5 & 7 \\
   5 & 2 & 4 & 7 & 1 & 3 & 6 \\
   6 & 2 & 3 & 5 & 1 & 4 & 7 \\
   7 & 3 & 4 & 6 & 1 & 2 & 5 \\
   \end{array}
   \]

   This table satisfies the condition that for any two players, there is a third player who beats both.

The final answer is \( \boxed{ k = 7 } \).