For example, $3 n(\geqslant 5)$ teams participate in a round-robin tournament, where the winner gets 3 points, a draw results in each team getting 1 point, and the loser gets 0 points. The result is that the team ranked third from the bottom scores fewer points than the teams ahead of it but more than the last two teams; and its number of wins is more than the teams ahead of it but fewer than the last two teams. Find the minimum value of the number of teams $n$.

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Let the $n$ teams be arranged in order of their scores as $A_{1}, A_{2}, \cdots, A_{n}$. Suppose $A_{i}$ wins $m_{i}$ games and draws $k_{i}$ games, then the points are $3 m_{i} + k_{i}$. According to the problem, when $i=1,2$, $m_{i} \geqslant m_{3} + 1$ and $3 m_{i} + k_{i} \leqslant 3 m_{3} + k_{3} - 1$, and when $i \geqslant 4$, $m_{i} \leqslant m_{3} - 1$ and $3 m_{i} + k_{i} \geqslant 3 m_{3} + k_{3} + 1$. Therefore, the total number of games is
$$
\begin{array}{l}
\frac{1}{2} n(n-1) = C_{n}^{2} = \sum_{i=1}^{n} m_{i} + \frac{1}{2} \sum_{i=1}^{n} k_{i}. \\
\begin{aligned}
n(n-1) & = 2 \sum_{i=1}^{n} m_{i} + \sum_{i=1}^{n} k_{i} \\
& \geqslant 2 \sum_{i=1}^{2} m_{i} + 2 m_{3} + 2 \sum_{i=4}^{n} m_{i} + k_{3} + \sum_{i=4}^{m} k_{i} \\
& \geqslant 4(m_{3} + 1) + 2 m_{3} + 2 \sum_{i=4}^{n} m_{i} + k_{3} + \sum_{i=4}^{n} k_{i} \\
& = 6 m_{3} + 4 + k_{3} + \sum_{i=4}^{n} (3 m_{i} + k_{i}) - \sum_{i=4}^{n} m_{i} \\
& \geqslant 6 m_{3} + 4 + k_{3} + (n-3)(3 m_{3} + k_{3} + 1) - (n-3)(m_{3} - 1) \\
& = 2 n m_{3} + (n-2) k_{3} + 2 n - 2.
\end{aligned}
\end{array}
$$

From the condition that when $i \geqslant 4$, $0 \leqslant m_{i} \leqslant m_{3} - 1$, we know $m_{3} \geqslant 1$. Therefore, when $i=1,2$, $m_{i} \geqslant m_{3} + 1 \geqslant 2$, meaning $A_{1}$ and $A_{2}$ each win at least two games. To have a higher score than $A_{1}$ and $A_{2}$, $A_{3}$ must draw at least 4 more games than each of $A_{1}$ and $A_{2}$, so $k_{3} \geqslant 4$. Similarly, when $i \geqslant 4$, $A_{i}$ must draw at least 4 more games than $A_{3}$, meaning each $A_{i} (i \geqslant 4)$ must draw at least 8 games, so $n \geqslant 9$. Hence, $A_{3}$ must have at least 1 win, 4 draws, and 3 losses, and $A_{1}$ and $A_{2}$ each must have at least 2 wins and 6 losses. Therefore, $A_{1}$, $A_{2}$, and $A_{3}$ together must lose at least $3 + 6 + 6 = 15$ games. Excluding the 3 games they play against each other, they must lose at least 12 games to $A_{4}, A_{5}, \cdots, A_{n}$. Thus, $A_{4}, A_{5}, \cdots, A_{n}$ must win at least 12 games against $A_{1}, A_{2}, A_{3}$. If each of $A_{4}, A_{5}, \cdots, A_{n}$ wins only one game, then $n-3 \geqslant 12$, so $n \geqslant 15$. Otherwise, at least one of $A_{4}, A_{5}, \cdots, A_{n}$, say $A_{i_{0}} (1 \leqslant i_{0} \leqslant n)$, wins 2 games, i.e., $m_{i_{0}} \geqslant 2$, so $m_{3} \geqslant m_{i_{0}} + 1 \geqslant 3$. Since $k_{3} \geqslant 4$, substituting into (1) gives
$$
n^{2} - n \geqslant 6 n + 4(n-2) + 2 n - 2,
$$
i.e., $n^{2} - 13 n + 10 \geqslant 0$.
Thus, $n \geqslant \frac{13 + \sqrt{13^{2} - 4 \times 10}}{2} = \frac{13 + \sqrt{129}}{2} > \frac{13 + 11}{2} = 12$.

Therefore, $n \geqslant 13$. When $n=13$, $A_{3}$ has 3 wins, 4 draws, and 5 losses, $A_{1}$ and $A_{2}$ each have 4 wins and 8 losses, and $A_{4}, A_{5}, \cdots, A_{13}$ each have 2 wins, 9 draws, and 1 loss. As shown in the table, this result is achievable.
Thus, the minimum value of $n$ is 13.