18. (16th All-Russian Mathematics Competition) In the school football championship, each team must play a match against every other team. Each match awards 2 points to the winning team, 1 point to each team in the event of a draw, and 0 points to the losing team. It is known that one team has the highest score, but it has won fewer matches than any other team. How many teams must have participated at a minimum?
Let the team with the highest score, denoted as team $A$, be the champion. 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 win at least $n + 1$ matches, meaning their score is no less than $2(n + 1)$ points. Therefore,
$$
\begin{array}{l}
2n + m > 2(n + 1), \\
m \geq 3.
\end{array}
$$

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

Let there be $s$ teams in total. The champion must win at least one match; otherwise, its score would not exceed $s - 1$ points. Any other team would score strictly less than $s - 1$ points, and the total score of all participating teams would be less than $s(s - 1)$ points, while the total score of $s$ teams is $s(s - 1)$ points, leading to a contradiction.

Thus, $m > 4, n \geq 1$, meaning the champion team $A$ must play at least 5 matches, implying there are at least 6 teams in the competition.

A score table for 6 teams can always be constructed to meet the given conditions, where the champion team $A$ wins the minimum number of matches:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & Score \\
\hline$A$ & & 1 & 1 & 1 & 1 & 2 & 6 \\
\hline$B$ & 1 & & 2 & 0 & 0 & 2 & 5 \\
\hline$C$ & 1 & 0 & & 0 & 2 & 2 & 5 \\
\hline$D$ & 1 & 2 & 2 & & 0 & 0 & 5 \\
\hline$E$ & 1 & 2 & 0 & 2 & & 0 & 5 \\
\hline$F$ & 0 & 0 & 0 & 2 & 2 & & 4 \\
\hline
\end{tabular}