For example, 330 football teams, each team has played the same number of matches, and each match has a winner (no draws). What is the maximum number of teams that can have more wins than losses?

Each team has played the same number of matches, and each match has a winner (no draws). The number of teams with more wins than losses is at most how many?
In each match, there is one win and one loss, so the total number of wins by all teams is exactly equal to the total number of losses by all teams.

If each team has more wins than losses, then the total number of wins by all teams would be greater than the total number of losses by all teams, which is a contradiction. Therefore, at most 29 teams can have more wins than losses.
We show that it is possible for 29 teams to have more wins than losses. For this,
represent these 29 teams with 29 points, denoted as $v_{1}, v_{2}, \cdots, v_{29}$. Let $v_{r+2 s}=v_{x}(i=1,2, \cdots)$.
Let $v_{1}$ win against $v_{r+1}, v_{i+2}, \cdots, v_{i+14}(i=1,2, \cdots)$, then each of these 29 teams wins 14 matches and loses 14 matches. Add another point $u$ representing the 30th team, which loses to $v_{1}, v_{2}, \cdots, v_{29}$, then $v_{1}, v_{2}, \cdots, v_{29}$ all have more wins than losses.
Note 1: If a directed edge is drawn from the winning team to the losing team, then there is a directed edge between every pair of the 30 points. Such a graph is called a tournament graph. If an edge (not a directed edge) is drawn between every pair of points, such a graph is called a complete graph. See Example in Section 3.7.
Note 2: More generally, if 30 is replaced by $n(\geqslant 2)$, then the number of teams with more wins than losses is at most: