Example 3 Before the World Cup, a coach of a certain country, in order to examine the players $A_{1}, A_{2}, \cdots, A_{7}$, plans to have them all play in three training matches (each 90 minutes). Assuming that at any moment during the matches, only 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, then, based on the total playing time of each player, how many different scenarios are there?
Let $A_{i}(i=1,2, \cdots, 7)$ be the total time in minutes that each player is on the field. According to the problem, we can set
$$
a_{i}=7 k_{i}(i=1,2,3,4), a_{i}=13 k_{i}(i=5,6,7),
$$

where $k_{i}(i=1,2, \cdots, 7) \in \mathbf{Z}_{+}$.
$$
\text { Let } \sum_{i=1}^{4} k_{i}=m, \sum_{i=5}^{7} k_{i}=n \text {, where } m \geqslant 4, n \geqslant
$$
3 , and $m, n \in \mathbf{Z}_{+}$, then $7 m+13 n=270$. It is easy to find an integer particular solution as $m=33, n=3$. Since $(7,13)=1$, the general integer solution is $m=33+13 t, n=3-7 t$.
Also, because $\left\{\begin{array}{l}33+13 t \geqslant 4, \\ 3-7 t \geqslant 3,\end{array}\right.$ we have $-\frac{29}{13} \leqslant t \leqslant 0$.
Thus, the integer $t=0,-1,-2$.
Therefore, all integer solutions that satisfy the conditions are
$$
(m, n)=(33,3),(20,10),(7,17) \text {. }
$$

For $\sum_{i=1}^{4} k_{i}=33$, the positive integer solutions can be written as a row of 33 ones, and in the 32 spaces between each pair of ones, choose 3 to insert “+” signs. Then, count the ones in the 4 parts separated by the 3 “+” signs to get one positive integer solution, so $\sum_{i=1}^{4} k_{i}=33$ has $\mathrm{C}_{32}^{3}$ positive integer solutions $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$;

Similarly, $\sum_{i=5}^{7} k_{i}=3$ has $\mathrm{C}_{2}^{2}$ positive integer solutions $\left(k_{5}, k_{6}, k_{7}\right)$.

Thus, the number of positive integer solutions $\left(k_{1}, k_{2}, k_{3}, k_{4}, k_{5}, k_{6}, k_{7}\right)$ that satisfy the conditions is $\mathrm{C}_{32}^{3} \cdot \mathrm{C}_{2}^{2}$...

Therefore, the total number of positive integer solutions $\left(k_{1}, k_{2}, k_{3}, k_{4}, k_{5}, k_{6}, k_{7}\right)$ that satisfy the conditions is
$$
\begin{array}{l}
\mathrm{C}_{32}^{3} \cdot \mathrm{C}_{2}^{2}+\mathrm{C}_{19}^{3} \cdot \mathrm{C}_{9}^{2}+\mathrm{C}_{6}^{3} \cdot \mathrm{C}_{16}^{2} \\
=4960+34884+2400=42244 \text { solutions. }
\end{array}
$$

Hence, based on the total time each player is on the field, there are 42244 different scenarios.