6.58 $n(\geqslant a)$ football teams participate in a round-robin tournament, and the team ranked third from the bottom scores fewer points than the teams ahead of it and more points than the two teams behind it. Additionally, this team has more wins than the teams ahead of it but fewer wins than the two teams behind it. Find the minimum value of the number of teams $n$.

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[Solution] Let the $n$ teams be $A_{1}, A_{2}, \cdots, A_{n-3}, B, C_{1}, C_{2}$, where team B is the third from the bottom, and $C_{1}$ and $C_{2}$ are the last two.

Firstly, note that each team in group A scores more than team B, and team B scores more than each team in group C; each team in group A has fewer wins than team B, and team B has fewer wins than each team in group C. Therefore, each team in group A must score at least 2 points more and have at least 2 fewer wins than any team in group C, which means each team in group A must have at least 8 draws, each team in group C must have at least 6 losses, and team B must have at least 3 losses, 4 draws, and 1 win. Thus, the last 3 teams must have at least 15 losses in total. The last 3 teams play 3 matches among themselves, with at most 3 losses, so the teams in group A must have at least 12 wins. If $n \leqslant 14$, then the number of teams in group A, $n-3 \leqslant 11$, and by the pigeonhole principle, there must be a team in group A with at least 2 wins and 8 draws, hence $n \geqslant 11$.

Assume $n=11$, then group A has 8 teams. The 8 teams in group A play $C_{8}^{2}=28$ matches among themselves. Even if all matches end in draws, there are only 56 draws within the group. However, it has been pointed out that each team in group A must have at least 8 draws, so the 8 teams in group A must have at least 64 draws in total. The 8 teams in group A must draw at least 8 matches with the last 3 teams. By the pigeonhole principle, either team B has 6 draws or a team in group C has 2 draws, which would lead to a team in group A having 2 wins and 10 draws, contradicting $n=11$.

Assume $n=12$, then group A has 9 teams. From the first argument, a team in group A must have 2 wins. Therefore, each team in group A must have 2 wins. Otherwise, a team in group A with 1 win must have 3 more draws, leading to a contradiction. If each team in group A has 2 wins and 8 draws, team B has 3 wins, 4 draws, and 3 losses, and each team in group C has 4 wins and 6 losses, then there are a total of 29 wins and 15 losses. At this point, each team plays 11 matches, and each team in group C has at most 7 losses, and team B has at most 4 losses, so the last 3 teams have at most 18 losses. Thus, the teams in group A must have at least 11 losses. By the pigeonhole principle, a team in group A must have 2 losses. Therefore, this team must have 2 wins, 8 draws, and 2 losses, totaling 12 matches, which is impossible.

When $n=13$, group A has 10 teams. If each team in group A has 2 wins, 9 draws, and 1 loss, team B has 3 wins, 4 draws, and 5 losses, and each team in group C has 4 wins and 8 losses, then the requirements are exactly met. The specific situation is shown in the table below, where all empty cells are filled with 1. It is easy to see that the results shown in the table fully meet the requirements of the problem.
In summary, the minimum value of the number of teams $n$ is 13.