On a tournament with $ n \ge 3$ participants, every two participants played exactly one match and there were no draws. A three-element set of participants is called a [i]draw-triple[/i] if they can be enumerated so that the ﬁrst defeated the second, the second defeated the third, and the third defeated the ﬁrst. Determine the largest possible number of draw-triples on such a tournament.
1. **Understanding the Problem:**
   We need to determine the largest possible number of draw-triples in a tournament with \( n \geq 3 \) participants, where each participant plays exactly one match against every other participant, and there are no draws. A draw-triple is a set of three participants \( \{a, b, c\} \) such that \( a \) defeats \( b \), \( b \) defeats \( c \), and \( c \) defeats \( a \).

2. **Characterizing Non-Draw-Triples:**
   A set \( \{a, b, c\} \) is not a draw-triple if and only if there is one participant who either defeats the other two or loses to the other two. This means that for each participant \( x_i \), we can pair them with two other participants \( y_i \) and \( w_i \) such that \( x_i \) beats \( y_i \) and loses to \( w_i \).

3. **Counting Non-Draw-Triples:**
   For each participant \( x_i \), let \( y_i \) be the number of participants \( x_i \) defeats and \( w_i \) be the number of participants \( x_i \) loses to. Clearly, \( y_i + w_i = n - 1 \).

4. **Calculating the Number of Draw-Triples:**
   The total number of ways to choose 3 participants out of \( n \) is \( \binom{n}{3} \). The number of non-draw-triples can be calculated by considering the pairs \( (y_i, w_i) \) for each participant \( x_i \). The number of non-draw-triples involving \( x_i \) is given by \( \frac{\binom{w_i}{2} + \binom{y_i}{2}}{2} \).

5. **Using Inequality to Find Upper Bound:**
   Using the inequality \( 2(w_i^2 + y_i^2) \geq (n - 1)^2 \), we can derive an upper bound for the number of draw-triples. This inequality helps us to bound the sum of the squares of \( w_i \) and \( y_i \).

6. **Calculating the Upper Bound:**
   \[
   \sum_{i=1}^{n} \left( \frac{\binom{w_i}{2} + \binom{y_i}{2}}{2} \right) \leq \frac{n^3 - 4n^2 + 3n}{8}
   \]
   Therefore, the maximum number of draw-triples is:
   \[
   \binom{n}{3} - \frac{n^3 - 4n^2 + 3n}{8}
   \]

7. **Checking for Equality:**
   - If \( n \) is odd, we can set \( y_i = w_i = \frac{n-1}{2} \) for each participant to achieve the upper bound.
   - If \( n \) is even, we can set \( w_i = \frac{n}{2} \) and \( y_i = \frac{n-2}{2} \) for each participant to achieve the upper bound.

8. **Final Expression for Even \( n \):**
   For even \( n \), the maximum number of draw-triples is:
   \[
   \binom{n}{3} - \frac{n^3 - 4n^2 + 4n}{8}
   \]

The final answer is \( \boxed{ \binom{n}{3} - \frac{n^3 - 4n^2 + 3n}{8} } \) for odd \( n \) and \(\binom{n}{3} - \frac{n^3 - 4n^2 + 4n}{8}\) for even \( n \).