For example, 312 players participate in a table tennis round-robin tournament, where each pair of players competes in one match. If player $A_{i}$ beats $A_{j}$, $A_{j}$ beats $A_{k}$, and $A_{k}$ beats $A_{i}$, then $\left\{A_{i}, A_{j}, A_{k}\right\}$ is called a triangle. Let $L_{i}, W_{i}$ represent the number of wins and losses for player $A_{i}$. It is known that if $A_{i}$ beats $A_{j}$, then $L_{i} + W_{j} \geqslant 10$. Try to find the number of triangles in this round-robin tournament.
Given the problem, $\sum_{i=1}^{12} W_{i}=\sum_{i=1}^{12} L_{i}=C_{12}^{2}=66$, and for any $i(1 \leqslant i \leqslant 12), L_{i}+W_{i}=11$. Without loss of generality, let $W_{1}=\max _{1 \leqslant i \leqslant 12}\left\{W_{i}\right\}$, then $W_{1} \geqslant \frac{1}{12} \sum_{i=1}^{12} W_{i}=5.5$, so $\bar{w}_{1} \geqslant 6$.

If $W_{1} \geqslant 7$, then $L_{1}=11-W_{1} \leqslant 4$. Without loss of generality, let $A_{1}$ win against $A_{2}, A_{3}, \cdots, A_{8}$, then by $L_{1}+W_{i} \geqslant 10$, we get $W_{i} \geqslant 10-L_{1} \geqslant 6(i=2,3, \cdots, 8)$. Thus, $L_{i}=11-W_{i} \leqslant 5(i=2,3, \cdots, 8)$.

Consider the $C_{7}^{2}=21$ matches among $A_{2}, A_{3}, \cdots, A_{8}$. Without loss of generality, let $A_{8}$ win the fewest matches, so $A_{8}$ wins $\leqslant\left[\frac{21}{7}\right]=3$ matches. But $W_{8} \geqslant 6$. Therefore, $A_{8}$ must win 3 matches against $A_{9}, A_{10}, A_{11}, A_{12}$. Without loss of generality, let $A_{8}$ win against $A_{9}$, $A_{10}, A_{11}$. By $L_{8}+W_{j} \geqslant 10$ and $L_{8} \leqslant 11-W_{8} \leqslant 5$, we get $W_{j} \geqslant 10-L_{8} \geqslant 5(j=9,10,11)$, thus $L_{j}=11-W_{j} \leqslant 6(j=9,10,11)$.

If one of $A_{1}, A_{2}, \cdots, A_{11}$, say $A_{i_{0}}$, wins against $A_{12}$, then $L_{i_{0}}+W_{12} \geqslant 10, W_{12} \geqslant 10-L_{i_{0}} \geqslant 4$; if $A_{1}$, $A_{2}, \cdots, A_{11}$ all lose to $A_{12}$, then $W_{12} \geqslant 11$. Hence, $W_{12} \geqslant 4$. Therefore,
$\sum_{i=1}^{12} W_{i} \geqslant 7+7 \times 6+3 \times 5+4=68>66$, a contradiction.

Thus, $W_{1}=6$, and hence $W_{i} \leqslant W_{1}=6(1 \leqslant i \leqslant 12)$. Similarly, we can prove $L_{i} \leqslant 6(1 \leqslant i \leqslant 12)$, and $L_{i}+W_{i}=11$. Therefore, $L_{i}, W_{i}$ can only be 5 or 6. Let $W_{1}, W_{2}, \cdots, W_{12}$ have $x$ values equal to 5 and $y$ values equal to 6, then
$$
\left\{\begin{array}{l}
5 x + 6 y = 66, \\
x + y = 12,
\end{array}\right. \text{ so } \left\{\begin{array}{l}
x=6, \\
y=6.
\end{array}\right.
$$

Without loss of generality, let $W_{1}=W_{2}=\cdots=W_{6}=5, W_{7}=W_{8}=\cdots=W_{12}=6$, then $L_{1}=L_{2}=\cdots=L_{6}=6, L_{7}=$ $L_{8}=\cdots=L_{12}=5$.

Use 12 points in a plane (still denoted as $A_{1}, A_{2}, \cdots, A_{12}$) to represent the 12 players. If $A_{i}$ wins against $A_{j}$, draw an arrow from $A_{i}$ to $A_{j}$: $A_{i} \rightarrow A_{j}$, to get a directed graph $G$. The graph has $C_{12}^{2}=220$ triangles, which are divided into two types: in the (I) type triangles, each vertex has two arrows, one incoming and one outgoing. The rest of the triangles belong to the (II) type (Figure 1-14). The number of triangles we are looking for is the number of (I) type triangles $u$. Let the number of (II) type triangles be $v$, then
$$
u+v=220.
$$

We call the angle formed by two arrows from a point a "shooting angle". On one hand, when $1 \leqslant i \leqslant 6$, the number of shooting angles with $A_{i}$ as the vertex is $C_{5}^{2}$, and when $7 \leqslant i \leqslant 12$, the number of shooting angles with $A_{i}$ as the vertex is $C_{6}^{2}$. Therefore, the total number of shooting angles in graph $G$ is
$$
S=6\left(C_{5}^{2}+C_{6}^{2}\right)=150.
$$

On the other hand, in (I) type triangles, there are no shooting angles, and in (II) type triangles, there is exactly one shooting angle. Therefore, $S=v$, so $v=150$. From (1), we get
$$
u=220-v=70.
$$

Thus, the number of triangles we are looking for is 70.