17. 14 people participate in a Japanese chess round-robin tournament, where each person plays against the other 13 people, and there are no draws in the matches. Find the maximum number of "triangles" (here, a "triangle" refers to a set of three people where each person has one win and one loss).

(2002, Japan Mathematical Olympiad (First Round))
Solution: Let the 14 people be represented as $A_{1}, A_{2}, \cdots, A_{14}$, and let $A_{4}$ win $a_{4}$ games. In any group of three people (i.e., selecting 3 out of 14 people), if there is no "double corner," then one person must have won the other two. Therefore, the number of groups of three that do not form a "triple corner" is $\sum_{k=1}^{14} \mathrm{C}_{a_{k}}^{2}$ (here, define $C_{1}^{2}=C_{0}^{2}=0$). The number of "triple corners" is $C_{14}^{3}-$ $\sum_{k=1}^{14} C_{a_{k}}^{2}$, which means we need to find the minimum value of $\sum_{k=1}^{14} C_{a_{k}}^{2}$.
It is known that $\sum_{k=1}^{14} a_{k}=C_{14}^{2}=91$. Below, we prove:
(1) For integers $x_{k} \geqslant 0$, and satisfying $\sum_{i=1}^{14} x_{k}=91$. The minimum value of $\sum_{k=1}^{14} \mathrm{C}_{x_{k}}^{2}$ is 252;
(2) There exists a win-loss scenario that satisfies $\sum_{i=1}^{14} \mathrm{C}_{a_{k}}^{2}=252$.
First, we prove (1). Since each $x_{i}$ has only a finite number of possible values, the minimum value of $\sum_{k=1}^{14} \mathrm{C}_{x_{k}}^{2}$ exists. Suppose there exist $1 \leqslant i, j \leqslant 14$ such that $x_{1}-x_{j} \geqslant 2$. Define
$$
y_{k}=\left\{\begin{array}{ll}
x_{i}-1 & (k=i), \\
x_{j}+1 & (k=j), \\
x_{k} & (k \neq i, j) .
\end{array}\right.
$$

Then $y_{k} \geqslant 0, \sum_{k=1}^{14} y_{k}=91$, and
$$
\begin{array}{l} 
\sum_{k=1}^{14} \mathrm{C}_{x_{k}}^{2}-\sum_{k=1}^{14} \mathrm{C}_{y_{k}}^{2}=\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{j}}^{2}-\mathrm{C}_{x_{i}-1}^{2}-\mathrm{C}_{x_{j}+1}^{2} \\
= \frac{x_{1}\left(x_{i}-1\right)}{2}+\frac{x_{j}\left(x_{j}-1\right)}{2}-\frac{\left(x_{i}-1\right)\left(x_{i}-2\right)}{2}- \\
\frac{\left(x_{j}+1\right) x_{j}}{2} \\
= x_{i}-x_{j}-1>0 .
\end{array}
$$

Therefore, $\sum_{k=1}^{14} C_{x_{k}}^{2}$ is not the minimum value. Thus, the $x_{k}$ that makes $\sum_{k=1}^{14} C_{x_{k}}^{2}$ reach its minimum value should satisfy: for all $1 \leqslant i, j \leqslant 14,\left|x_{1}-x_{j}\right| \leqslant 1$. Clearly,
$$
\begin{array}{l}
\left\{x_{1}, x_{2}, \cdots, x_{14}\right\} \\
=\{6,6,6,6,6,6,6,7,7,7,7,7,7,7\} .
\end{array}
$$

Thus, the minimum value of $\sum_{k=1}^{14} \mathrm{C}_{\mathrm{r}_{k}}^{2}$ is $\mathrm{C}_{6}^{2} \times 7+\mathrm{C}_{7}^{2} \times 7=252$.
Next, we prove (2).
The win-loss situation in each cycle is as shown in Table 2.
Here, in the $(i, j)$ table, “O” indicates $A_{1}$ wins $A_{j}$, “$\bullet$” indicates $A_{i}$ loses to $A_{j}$, and “$\times$” indicates that no match was played between $A_{i}$ and $A_{j}$.
Therefore, the maximum number of "double corners" is
$$
C_{14}^{3}-252=112 .
$$