Example 22 (2002 National High School Mathematics Competition Question) Before the World Cup, Coach of Country $F$ is evaluating seven players, $A_{1}, A_{2}, \cdots, A_{7}$, and plans to have them all play in three training matches (each 90 minutes long). Assuming that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in minutes) of $A_{1}, A_{2}, A_{3}, A_{4}$ is divisible 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?
Let the playing time of the $i$-th player be $x_{i}$ minutes $(i=1,2, \cdots, 7)$. The problem is transformed into finding the number of positive integer solutions to the indeterminate equation $x_{1}+x_{2}+\cdots+x_{7}=270$
(*)
satisfying the conditions $7 \mid x_{i}(i=1,2,3,4)$, and $13 \mid x_{j}(i=5,6,7)$. 
Let $x_{1}+x_{2}+x_{3}+x_{4}=7 m, x_{5}+x_{6}+x_{7}=13 n$, then
$7 m+13 n=270$, and $m, n \in \mathbf{N}^{+}, m \geqslant 4, n \geqslant 3$.
It is easy to find that the positive integer solutions $(m, n)$ satisfying the above conditions are
$$
(m, n)=(33,3),(20,10),(7,17) \text {. }
$$

When $(m, n)=(33,3)$, $x_{5}=x_{6}=x_{7}=13$. Let $x_{i}=7 y_{i}(i=1,2,3,4)$, then $y_{1}+y_{2}+$ $y_{3}+y_{4}=33$, there are $\mathrm{C}_{33-1}^{4-1}=\mathrm{C}_{32}^{3}=4960$ groups of positive integer solutions $\left(y_{1}, y_{2}, y_{3}, y_{4}\right)$, at this time there are 4960 groups of positive integer solutions satisfying the conditions.

When $(m, n)=(20,10)$, let $x_{i}=7 y_{i}(i=1,2,3,4), x_{j}=13 y_{j}(j=5,6,7)$, then $y_{1}+y_{2}+y_{3}+y_{4}=20, y_{5}+y_{6}+y_{7}=10$, at this time (*) has $\mathrm{C}_{19}^{3} \cdot \mathrm{C}_{9}^{2}=34884$ groups of solutions.

When $(m, n)=(7,17)$, let $x_{i}=7 y_{i}(i=1,2,3,4), x_{j}=13 y_{j}(j=5,6,7)$, then $y_{1}+$ $y_{2}+y_{3}+y_{4}=7, y_{5}+y_{6}+y_{7}=17$, at this time ( *) has $\mathrm{C}_{6}^{3} \cdot \mathrm{C}_{16}^{2}=2400$ groups of solutions.
Therefore, the number of positive integer solutions satisfying $(*)$ is $4960+34884+2400=42244$ groups.