18.14 In a Japanese chess round-robin tournament, each of the 14 participants plays against the other 13. There are no ties in the matches. Find the maximum number of "triangular ties" (here, a "triangular tie" refers to a situation where each of the 3 participants has one win and one loss against the other two).
18. Let $A_{1}, A_{2}, \cdots, A_{14}$ represent 14 people, and let $A_{i}$ win $a_{i}$ games $(i=1,2, \cdots, 14)$. If any 3 people do not form a "triangle link," then one of them must have won the other two, so the number of "3-person groups" that do not form a "triangle link" is $\sum_{i=1}^{14} C_{a_{i}}^{2}$ (with the convention that $C_{0}^{2}=C_{1}^{2}=0$). Therefore, the total number of "triangle links" is $S=C_{14}^{3}-\sum_{i=1}^{14} C_{a_{i}}^{2}$. Using the adjustment method, it is easy to prove that when $\sum_{i=1}^{14} C_{a_{i}}^{2}$ takes its minimum value, there must be $\left|a_{i}-a_{j}\right| \leqslant 1(1 \leqslant i<j \leqslant 14)$. Also, $\sum_{i=1}^{14} a_{i}=C_{14}^{2}=91$, from which it follows that $a_{1}, a_{2}, \cdots, a_{14}$ includes 7 instances of 6 and 7 instances of 7. Hence, the minimum value of $\sum_{i=1}^{14} C_{a_{i}}^{2}$ is $7 \times C_{7}^{2}+7 \times C_{6}^{2}=252$, so the maximum value of $S=C_{14}^{3}-\sum_{i=1}^{14} C_{a_{i}}^{2}$ is $C_{14}^{3}-252=112$. On the other hand, when $1 \leqslant i \leqslant 14$, let $A_{i}$ win against $A_{i+1}, A_{i+2}, \cdots, A_{i+6}$ (with the convention that $A_{j+14}=A_{j}$) these 6 teams and lose to $A_{i+7}, A_{i+8}, \cdots, A_{i+13}$ (with the convention that $A_{j+14}=A_{j}$) these 7 teams. When $8 \leqslant i \leqslant 14$, let $A_{i}$ win against $A_{i+1}, A_{i+2}, \cdots, A_{i+7}$ (with the convention that $A_{j+14}=A_{j}$) these 7 teams and lose to $A_{i+8}, A_{i+9}, \cdots, A_{i+13}$ (with the convention that $A_{j+14}=A_{j}$) these 6 teams. This arrangement ensures that $\sum_{i=1}^{14} C_{a_{i}}^{2}=252$ holds. In summary, the maximum number of "triangle links" is 112.