Example 5 In the school football championship, it is required that each team must play a match against all the other teams. Each winning team gets 2 points, a draw gives each team 1 point, and a losing team gets 0 points. It is known that one team scored the most points, but it played fewer matches than any other team. How many teams participated at least?
Explanation: We call the team $A$ with the highest score the winning team.
Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.

From the given conditions, every other team must play at least $n+1$ matches, meaning they score no less than $2(n+1)$ points. Therefore,
$2n + m > 2(n+1)$. This implies $m \geqslant 3$.
Thus, there must be a team that draws with the winning team $A$, and this team's score should be no less than $2(n+1) + 1$ points, i.e.,
$$
2n + m > 2(n+1) + 1, \quad m \geqslant 4.
$$

Let there be $x$ teams in total, then the winner must win at least one match. Otherwise, its score would not exceed $x-1$ points. In this case, any other team's score would be strictly less than $x-1$ points, and the total score of all participating teams would be less than $x(x-1)$ points, but the total score of $x$ participating teams is $x(x-1)$ points, leading to a contradiction.

Thus, $m \geqslant 4, n \geqslant 1$, meaning the winning team $A$ must play at least 5 matches, and therefore, there must be at least 6 teams participating.

From this, we can estimate: at least 6 teams are participating in the competition.

We can construct a match points table: Table 1 shows 6 teams (let's call them $A, B, C, D, E, F$) participating, with the winning team $A$ having the highest score and the fewest draws.
Table 1 Match Points Table
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & Score \\
\hline 4 & lle & 1 & 1 & 1 & 1 & 2 & 6 \\
\hline$B$ & 1 & lle & 2 & 0 & 0 & 2 & 5 \\
\hline i & 1 & 0 & Illo & 0 & 2 & 2 & 5 \\
\hline$D$ & 1 & 2 & 2 & & 0 & 0 & 5 \\
\hline$E$ & 1 & 2 & 0 & 2 & & 0 & 5 \\
\hline $\boldsymbol{r}$ & 0 & 0 & 0 & 2 & 2 & 16 & 4 \\
\hline
\end{tabular}