II. $n(\geqslant 5)$ football teams participate in a round-robin tournament. Each pair of teams plays one match, with the winning team getting 3 points, the losing team getting 0 points, and each team getting 1 point in the case of a draw. The team that finishes in 3rd to last place scores fewer points than all the teams above it, but more than the last two teams; it also plays more matches than the teams above it, but fewer than the last two teams. Find the minimum value of the number of teams $n$.

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Solution: Let the $n$ teams be $A_{1}, A_{2}, \cdots, A_{n-3}, B, C_{1}$, $C_{2}$.

Since the $A$ group teams score more but have fewer wins, and the $C$ group teams score less but have more wins, each team in the $A$ group must have at least 8 draws, team $B$ must have at least 1 win, 3 losses, and 4 draws, and each team in the $C$ group must have at least 2 wins and 6 losses. The last 3 teams must have at least 15 losses, and the matches among these 3 teams can be at most 3. Therefore, each team in the $A$ group must have at least 12 wins.

If $n \leqslant 14$, then the number of teams in the $A$ group $\leqslant 11$, which means some team in the $A$ group must have 2 wins and 8 draws, so $n \geqslant 11$.

Assume $n=11$. Then, the $A$ group has 8 teams, each with at least 8 draws, totaling 64 draws. The 8 teams in the $A$ group play $C_{8}^{2}=28$ matches. Each team in the $A$ group must draw at least 8 matches, including at least 4 draws with the $C$ group. Each team in the $C$ group must have at least 2 draws. Thus, each team in the $A$ group must have at least 2 wins and 10 draws, leading to a contradiction.

Assume $n=12$. Then, the $A$ group has 9 teams. Each team in the $C$ group must have at least 6 losses, and team $B$ must have 3 losses. The total number of losses for teams $B$ and $C$ is at least 15, so each team in the $A$ group must win at least 12 matches against teams in other groups. Thus, each team in the $A$ group must win at least 2 matches (if some teams in the $A$ group win 2 matches and others win 1 match, the teams that win 1 match must have 3 additional draws, leading to each team in the $C$ group having at least 9 losses and 4 wins, which contradicts $n=12$). Therefore, each team in the $C$ group must win at least 4 matches, and each team in the $A$ group must have some losses. This leads to each team in the $A$ group having 2 wins, 1 loss, and 8 draws or 2 wins and 9 draws, team $B$ having 3 wins, 4 losses, and 4 draws, and each team in the $C$ group having 4 wins and 7 losses. The total number of wins is $18+3+8=29$ matches, and the total number of losses is $\leqslant 9+4+14=27$, leading to a contradiction.

Assume $n=13$. The $A$ group has 10 teams, each with 2 wins, 1 loss, and 9 draws, team $B$ with 3 wins, 5 losses, and 4 draws, and each team in the $C$ group with 4 wins and 8 losses. The table is as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $A_{1}$ & $A_{2}$ & $A_{3}$ & $A_{4}$ & $A_{5}$ & $A_{6}$ & $A_{7}$ & $A_{8}$ & $A_{9}$ & $A_{10}$ & $B$ & $C_{1}$ & $C_{2}$ & $\cdots$ & Score \\
\hline$A_{1}$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $0: 3$ & $3: 0$ & $3: 0$ & 2 & 15 \\
\hline$A_{2}$ & $1: 1$ & & $1: 1$ & $0: 3$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & 2 & 15 \\
\hline$A_{3}$ & $1: 1$ & $1: 1$ & & $1: 1$ & $0: 3$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & 2 & 15 \\
\hline$A_{4}$ & $1: 1$ & $3: 0$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & 0.3 & $3: 0$ & 2 & 15 \\
\hline$A_{5}$ & $1: 1$ & $1: 1$ & $3: 0$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $0: 3$ & $3: 0$ & 2 & 15 \\
\hline$A_{6}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $0: 3$ & $3: 0$ & 2 & 15 \\
\hline$A_{7}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & $0: 3$ & 2 & 15 \\
\hline$A_{8}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & $0: 3$ & 2 & 15 \\
\hline$A_{4}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $3: 0$ & $3: 0$ & $0: 3$ & 2 & 15 \\
\hline$A_{10}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $3: 0$ & $3: 6$ & $0: 3$ & 2 & 15 \\
\hline$B$ & $3: 0$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & & $3: 0$ & $3: 0$ & 3 & 13 \\
\hline$C_{1}$ & $0: 3$ & $0: 3$ & $0: 3$ & $3: 0$ & $3: 0$ & $2: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & & $3: 0$ & 4 & 12 \\
\hline$C_{2}$ & $0: 3$ & $0: 3$ & $0: 2$ & $0: 3$ & $0: 3$ & $1: 3$ & $3: 0$ & $2: 0$ & $3: 0$ & $3: 0$ & $0: 3$ & $0: 3$ & & 4 & 12 \\
\hline
\end{tabular}

In summary, the minimum value of $n$ is 13.