Example 3: There are $n(n \geqslant 5)$ football teams participating 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 the third-to-last place scores fewer points than the teams ahead of it and more points than the last two teams; it also has more wins than the teams ahead of it but fewer wins than the last two teams. Find the minimum value of the number of teams $n$.

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Let $A_{1}, A_{2}, \cdots, A_{n-3}$ be the teams with high scores, $B$ be the team with the third-lowest score, and $C_{1}, C_{2}$ be the teams with the lowest scores. Introduce parameters: Let $B$ win $x$ games, draw $y$ games, and lose $z$ games, then $n=x+y+z+1$. Since $C_{1}$ wins at least $(x+1)$ games, we have $3 x+y \geqslant 3(x+1)+1$, i.e., $y \geqslant 4$. Since $A_{1}$ wins at most $(x-1)$ games and draws at most $(n-x)$ games, we have $3 x+y+1 \leqslant 3(x-1)+(n-x)=3 x+y+z-2$, thus $z \geqslant 3$, meaning $B$ loses at least 3 games. Therefore, at least one of $A_{1}, \cdots, A_{n-3}$ wins against $B$. Thus, $x \geqslant 2$.

(1) If all $A_{i}$ draw with at least one of $C_{1}, C_{2}, B$, since $y=n-x-z-1 \leqslant n-6$, at least 3 of the $(n-3)$ teams draw with one of $C_{1}, C_{2}$. Thus, one of $C_{1}, C_{2}$ draws at least two games, so $3 x+y \geqslant 3(x+1)+2+1$, i.e., $y \geqslant 6$. Thus, $n \geqslant 12$. However, the equality cannot hold, otherwise by $z=3$ we know that $A_{i}$ has not lost any games, and wins at most $x-1=1$ games, so they must draw at least 10 games. This implies that $B, C_{1}, C_{2}$ together draw at least 18 games, but since $y=6$, $C_{1}, C_{2}$ can draw at most 2 games, leading to $18 \leqslant 6+2+2$, a contradiction. Therefore, $n \geqslant 13$.

(2) If there is an $A_{i}$ (let it be $A_{1}$) that does not draw with $B, C_{1}, C_{2}$, then it draws at most $(n-4)$ games, so $3 x+y+1 \leqslant 3(x-1)+(n-4)$, giving $n \geqslant y+8 \geqslant 12$. The equality cannot hold either, otherwise $y=4$, and from the previous derivation, $C_{1}, C_{2}$ have not drawn any games, and $A_{i}$ wins at least $(x-1)$ games, otherwise it would draw 11 games, but $C_{1}$ has not drawn any games, a contradiction. Thus, $A_{i}$ wins $(x-1)$ games and draws 8 or 9 games. Suppose $k$ of $A_{i}$ draw 8 games, and $(9-k)$ draw 9 games, and $C_{1}, C_{2}$ each win $(x+1)$ games and lose $(10-x)$ games, we get $9(x-1)+x+2(x+1)=k(4-x)+(9-k)(3-x)+7-x+2(10-x)$, i.e., $24 x=k+61$. This equation has no integer solutions for $0 \leqslant k \leqslant 9$, so $n \geqslant 13$.

Combining (1) and (2), for all cases, we have $n \geqslant 13$.
On the other hand, when $n=13$, let the results of the matches between the teams be as shown in the table below, which shows that $n=13$ meets the requirements. Therefore, $n_{\text {min }}=13$.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $A_{1}$ & $A_{2}$ & $A_{3}$ & $A_{4}$ & $A_{5}$ & $A_{6}$ & $A_{7}$ & $A_{8}$ & $A_{9}$ & $A_{10}$ & $B$ & $C_{1}$ & $C_{2}$ \\
\hline$A_{1}$ & & 3 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 3 \\
\hline$A_{2}$ & 0 & & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 3 & 3 \\
\hline$A_{3}$ & 1 & 1 & & 3 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 3 \\
\hline$A_{4}$ & 1 & 1 & 0 & & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 3 & 3 \\
\hline$A_{5}$ & 1 & 1 & 1 & 1 & & 1 & 1 & 1 & 1 & 1 & 3 & 0 & 3 \\
\hline$A_{6}$ & 1 & 1 & 1 & 1 & 1 & & 1 & 1 & 1 & 1 & 3 & 3 & 0 \\
\hline$A_{7}$ & 1 & 1 & 1 & 1 & 1 & 1 & & 1 & 1 & 1 & 3 & 3 & 0 \\
\hline$A_{8}$ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & 1 & 1 & 3 & 3 & 0 \\
\hline$A_{9}$ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & 1 & 3 & 3 & 0 \\
\hline$A_{10}$ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & 0 & 3 & 3 \\
\hline$B$ & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 3 & & 3 & 3 \\
\hline$C_{1}$ & 3 & 0 & 3 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & & 3 \\
\hline$C_{2}$ & 0 & 0 & 0 & 0 & 0 & 3 & 3 & 3 & 3 & 0 & 0 & 0 & \\
\hline
\end{tabular}