11. In a city football tournament, each team must play one match against every other team, with 3 points awarded for a win, 1 point for a draw, and 0 points for a loss. It is known that one team has the highest score, more than any other team, but this team has the fewest wins, fewer than any other team. How many teams must there be at a minimum?
11. Let the team with the most points and the fewest wins be team $A$. If team $A$ wins $n$ games and draws $m$ games, then it scores $3n + m$ points. The other teams must win at least $n+1$ games, scoring at least $3(n+1)$ points. Therefore, $3(n+1) < 3n + m$, which implies $m \geq 4$. Team $A$ must draw at least 4 games. There is a team that draws with $A$, scoring at least $3(n+1) + 1$ points. From $3(n+1) + 1 < 3n + m$, we get $m \geq 5$. Let there be $k$ teams, then $k \geq m + 1 \geq 6$.
(1) If $k = 6$, team $A$ draws all 5 games, scoring 5 points. The other teams score at most 4 points each, so the total score is at most $5 + 4 \times 5 = 25$ points. With 6 teams, there are $\mathrm{C}_{6}^{2}$ games, and each game scores at least 2 points, so the total score is at least $\mathrm{C}_{6}^{2} \times 2 = 30$ points, which is a contradiction.
(2) If $k = 7$, since $m \geq 5$, team $A$ has three possibilities:
A. One win and five draws: Team $A$ scores 8 points, and the other 6 teams score at most 7 points each, so the total score is at most 50 points. The other teams win at least 2 games each, so the total number of wins is at least $1 + 2 \times 6 = 13$ games. The total score is at least $3 \times 13 + 2(\mathrm{C}_{7}^{2} - 13) = 55$ points, which is a contradiction.
B. Six draws: Team $A$ scores 6 points, and the other teams score at most 5 points each, so the total score is at most 36 points. The total score is at least $\mathrm{C}_{7}^{2} \times 2 = 42$ points, which is a contradiction.
C. One loss and five draws: Team $A$ scores 5 points, and the other teams score at most 4 points each, so the total score is at most 29 points. The total score is at least $\mathrm{C}_{7}^{2} \times 2 = 42$ points, which is a contradiction. Therefore, $k \geq 8$. When $k = 8$, the described situation can occur.
The match results are as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & Points & Wins \\
\hline$A$ & & 1 & 1 & 1 & 1 & 1 & 3 & 3 & 11 & 2 \\
\hline$B$ & 1 & & 3 & 0 & 3 & 0 & 3 & 0 & 10 & 3 \\
\hline$C$ & 1 & 0 & & 3 & 0 & 3 & 0 & 3 & 10 & 3 \\
\hline$D$ & 1 & 3 & 0 & & 3 & 0 & 3 & 0 & 10 & 3 \\
\hline$E$ & 1 & 0 & 3 & 0 & & 3 & 0 & 3 & 10 & 3 \\
\hline$F$ & 1 & 3 & 0 & 3 & 0 & & 3 & 0 & 10 & 3 \\
\hline$G$ & 0 & 0 & 3 & 0 & 3 & 0 & & 3 & 9 & 3 \\
\hline$H$ & 0 & 3 & 0 & 3 & 0 & 3 & 0 & & 9 & 3 \\
\hline
\end{tabular}
$$
k_{\text{min}} = 8.
$$