We consider sports tournaments with $n \ge 4$ participating teams and where every pair of teams plays against one another at most one time. We call such a tournament [i]balanced [/i] if any four participating teams play exactly three matches between themselves. So, not all teams play against one another.
Determine the largest value of $n$ for which a balanced tournament with $n$ teams exists.
1. **Lemma Statement and Proof:**
   We start by proving the lemma: If \( n \geq 5 \), there can't exist any triangle (three teams having met each other) or any antitriangle (three teams none of which faced any other).

   - **Proof of Lemma:**
     Assume, for contradiction, that there exists a triangle involving teams \( A, B, \) and \( C \). This means \( A \) has played against \( B \), \( B \) has played against \( C \), and \( C \) has played against \( A \). Now consider two other teams \( D \) and \( E \). By the hypothesis of the balanced tournament, neither \( D \) nor \( E \) has played against any of \( A, B, \) or \( C \). This leads to a contradiction because considering the four teams \( A, B, D, E \), there can be at most two matches among them, which violates the condition of having exactly three matches among any four teams. Therefore, no triangle can exist.

     Similarly, assume there exists an antitriangle involving teams \( A, B, \) and \( C \). This means none of \( A, B, \) and \( C \) have played against each other. Now consider two other teams \( D \) and \( E \). By the hypothesis, \( D \) and \( E \) must have played against each of \( A, B, \) and \( C \). This again leads to a contradiction because considering the four teams \( A, B, D, E \), there can be at most two matches among them, which violates the condition of having exactly three matches among any four teams. Therefore, no antitriangle can exist.

2. **Structure of the Tournament:**
   Given the lemma, we know that any four teams must be connected in a way such that they form a path (e.g., \( AB, BC, CD \)). Now consider five teams \( A, B, C, D, \) and \( E \). We can arrange them such that \( A, B, C, D \) form a path (in this order). Considering the teams \( B, C, D, E \), they must also be connected in that order to avoid creating a triangle or an antitriangle. Similarly, \( E \) must be connected to \( A \), creating a pentagon.

3. **Verification for \( n = 5 \):**
   The pentagon structure indeed works for \( n = 5 \). In a pentagon, each team plays against exactly two other teams, and any four teams among the five will have exactly three matches among them, satisfying the balanced tournament condition.

4. **Contradiction for \( n = 6 \):**
   Now assume there is a sixth team \( F \). By considering the teams \( A, B, C, \) and \( F \), there must be at least one edge of the form \( FA, FB, \) or \( FC \). This creates a contradiction because it would either form a triangle or an antitriangle, violating the balanced tournament condition.

Therefore, the maximum number of teams \( n \) for which a balanced tournament exists is \( n = 5 \).

The final answer is \( \boxed{5} \).