6.8 In the school football championship, each team must play a match against every other team. Each match's winning team gets 2 points, a draw gives each team 1 point, and the losing team gets 0 points. It is known that one team scored the most points (every other team scored fewer points than this team), but it won fewer matches than any other team. How many teams participated at the minimum?
［Solution］We call the team with the most points the winning team $A$.
Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given information, every other team must win at least $n+1$ matches, meaning their score is no less than $2(n+1)$ points. Therefore,
$$
\begin{array}{c}
2n + m > 2(n + 1), \\
m \geqslant 3.
\end{array}
$$

Thus, there must be a team that draws with team $A$, and this team's score should be no less than $2(n+1) + 1$ points. Therefore, we have
$$
\begin{array}{l}
2n + m > 2(n + 1) + 1, \\
m \geqslant 4.
\end{array}
$$

Let there be $s$ teams in total. The winner must win at least one match; otherwise, its score would not exceed $s-1$ points, and any other team's score would be strictly less than $s-1$ points. This would mean that the total score of all participating teams would be less than $s(s-1)$ points. However, the total score of $s$ participating teams is $2 C_{s}^{2} = s(s-1)$ points, leading to a contradiction.

Thus, $m \geqslant 4, n \geqslant 1$, meaning the winning team $A$ must play at least 5 matches, which implies there should be no fewer than 6 teams in the competition.

We can construct a match points table, which includes six teams $A, B, C, D, E, F$, and meets the requirements of the problem: the winning team $A$ wins the fewest matches but has the highest score.
In summary, there must be at least 6 teams participating.