Four, (18 points) In a round-robin football tournament with $n$ teams (i.e., 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 case of a draw, and 0 points to the losing team. The result is that one team has more points than any other team, yet has fewer wins than any other team. Find the smallest possible value of $n$.


Let the team with the highest score be Team $A$, and assume Team $A$ wins $k$ games, draws $m$ games, and thus Team $A$'s total points are $2k + m$.

From the given conditions, every other team must win at least $k+1$ games, and their points must be no less than $2(k+1)$. Therefore, we have $2k + m > 2(k+1)$, which simplifies to $m > 2$. Hence, $m \geqslant 3$.

Thus, there is a team that draws with Team $A$, and this team's score is no less than $2(k+1) + 1$. Therefore,
$$
2k + m > 2(k+1) + 1, \quad m > 3.
$$

This implies $m \geqslant 4$.
From this, we derive $n \geqslant m + 1 = 5$.
If $n = 5$, then $m = 4$, $k = 0$, and Team $A$ scores 4 points. However, in this case, 5 teams play a total of 10 games, and the total points of all teams sum up to 20 points. According to the problem, Team $A$'s score $>\frac{20}{5} = 4$, leading to a contradiction.
Therefore, $n \geqslant 6$.
When $n = 6$, there exists a match that meets the conditions (see the table below), so the minimum possible value of $n$ is 6.