Example 26 (2002 National High School League Additional Test Question 3, see Example 22 in Chapter 3) Before the World Cup, Coach of Country $F$ wants to evaluate seven players $A_{1}, A_{2}, \cdots, A_{7}$. He plans to have them all play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in minutes) of $A_{1}, A_{2}, A_{3}, A_{4}$ can all be divided by 7, while the total playing time (in minutes) of $A_{5}, A_{6}, A_{7}$ can all be divided by 13. If there are no limits on the number of substitutions per match, how many different scenarios are there for the total playing time of each player?
Decompose the construction of the indeterminate equation model. Let the playing time of the $i$-th player be $x_{i}(i=1,2, \cdots, 7)$. The problem is to find the number of positive integer solutions to the indeterminate equation $x_{1}+x_{2}+x_{3}+\cdots+x_{7}=270 \otimes$, under the conditions $7\left|x_{i}(1 \leqslant i \leqslant 4), 13\right| x_{i}(5 \leqslant i \leqslant 7)$. If $\left(x_{1}, x_{2}, \cdots, x_{7}\right)$ is a set of positive integer solutions that satisfy the conditions, then we have $\sum_{i=1}^{1} x_{i}=7 m, \sum_{i=5}^{7} x_{i}=13 n,\left(n, m \in \mathbf{N}^{*}\right)$, so $n, m$ are a set of positive integer solutions to the indeterminate equation $7 m+13 n=270,(m \geqslant 4, n \geqslant 3)$. By the theory of indeterminate equations, we can find that $7 m+13 n=270$, and the three sets of positive integer solutions that satisfy the conditions are
$$
\left\{\begin{array}{l}
m=33 \\
n=3
\end{array}, \quad\left\{\begin{array}{l}
m=20 \\
n=20
\end{array}, \quad\left\{\begin{array}{l}
m=7 \\
n=17
\end{array} .\right.\right.\right.
$$
1) When $m=33, n=3$, it is clear that $x_{5}=x_{6}=x_{7}=13$ has only one possibility; let $x_{i}=7 y_{i}(i=1,2,3,4)$, then the indeterminate equation $y_{1}+y_{2}+y_{3}+y_{4}=33$ has $C_{32}^{3}=4960$ sets of positive integer solutions, so $\otimes$ has 4960 sets of positive integers.
2) When $m=20, n=10$, similarly, we can find that the number of positive integer solutions to the original equation is $C_{19}^{3} \cdot C_{9}^{2}=34884$.
3) When $m=7, n=17$, there are $C_{6}^{3} \cdot C_{16}^{2}=2400$.

Thus, the number of sets of positive integer solutions to the original equation is $4960+34884+2400=42244$.