*5. In a football championship, each team is required to play a match against all other teams. Each match, the winning team gets 2 points, a draw results in 1 point for each team, and the losing team gets 0 points. It is known that one team has the highest score, but it has won fewer matches than any other team. If there are at least $n$ teams participating, then $n=$ $\qquad$ .
5. $n=6$. We call the team $A$ with the highest score the winning team. Suppose team $A$ wins $n$ games and draws $m$ games, then the total score of team $A$ is $2n + m$ points.

From the given conditions, each of the other teams must win at least $n+1$ games, i.e., score no less than $2(n+1)$ points. Therefore,
$2n + m > 2(n+1)$. This implies $m \geqslant 3$.
Thus, there exists a team that draws with team $A$ and scores no less than $2(n+1) + 1$, so we have
$2n + m > 2(n+1) + 1$, which implies $m \geqslant 4$.
Clearly, team $A$ must win at least one game, i.e., $n \geqslant 1$.
Therefore, team $A$ must play against at least 5 teams, meaning there are at least 6 teams participating in the competition. A tournament schedule for 6 teams that meets the conditions of the problem can be designed (omitted).