Example 3 A school is hosting a football round-robin tournament, where each participating team plays one match against every other team. A team earns 2 points for a win, 1 point for a draw, and 0 points for a loss. It is known that only one team has the highest points, but they have the fewest wins. How many teams must participate at minimum for this to be possible?
(1990 16th All-Russian Olympiad, Grade 9)
The team with the most points is called the champion. Suppose the champion wins $n$ games and draws $m$ games, then they have a total of $2n + m$ points. According to the assumption, the other teams win no less than $n+1$ games, meaning their points are no less than $2(n+1)$. From $2n + m > 2(n+1)$, we get $m \geqslant 3$, which means some teams have drawn, and their points are no less than $2(n+1) + 1$. From $2n + m > 2(n+1) + 1$, we get $m \geqslant 4$.

The champion team must win at least one game; otherwise, their points would not exceed $S-1$ (where $S$ is the number of participating teams), and the points of the other teams would all be less than $S-1$. Thus, the total points of all teams would be less than $S(S-1)$. However, for each game played, the total points of both teams combined is 2, so the total points of all teams should be $2 C_{s}^{2} = S(S-1)$, which is a contradiction. Therefore, $n \geqslant 1, m \geqslant 4$, meaning the champion team has participated in at least 5 games, and the number of participating teams (including the champion team) is no less than 6.

The following match points table shows that there are 6 teams (represented by $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$) participating, and the match results meet the requirements of the problem. Therefore, the minimum number of teams participating is 6.