3. Before the World Cup, the coach of country $F$ wants to examine $\mathrm{A}_{1}, \mathrm{~A}_{2}, \cdots, \mathrm{A}_{7}$, these seven players, and 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 $\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \mathrm{~A}_{4}$ is divisible by 7, and the total playing time (in minutes) of $\mathrm{A}_{5}, \mathrm{~A}_{6}, \mathrm{~A}_{7}$ 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.
3.【Analysis and Solution】Let the playing time of the $i$-th player be $x_{i}$ minutes $(i=1,2, \cdots, 7)$, the problem is to find the number of positive integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+\cdots+x_{7}=270
$$

under the conditions $7 \mid x_{i}(1 \leqslant i \leqslant 4)$ and $13 \mid x_{j}(5 \leqslant j \leqslant 7)$. If $\left(x_{1}, x_{2}, \cdots x_{7}\right)$ is a set of positive integer solutions satisfying condition (1), then $\sum_{i=1}^{4} X_{i}=7 m, \sum_{i=5}^{7} X_{i}=13 n, m, n \in \mathbf{N}_{+}$, so $m, n$ are a set of positive integer solutions to the indeterminate equation $7 m$ $+13 n=270 \quad(2)$ under the conditions $m \geqslant 4$ and $n \geqslant 3$.
Since $7(m-4)+13(n-3)=203$,
let $m^{\prime}=m-4, n^{\prime}=n-3$, we have $7 m^{\prime}+13 n^{\prime}=203$
Therefore, finding a set of positive integer solutions to (2) that satisfy the conditions $m \geqslant 4, n \geqslant 3$ is equivalent to finding a set of non-negative integer solutions to (3).
It is easy to observe that $7 \times 2+13 \times(-1)=1$, hence $7 \times 406+13 \times(-203)=203$.
That is, $m_{0}=406, n_{0}=-203$ is a particular integer solution to (3), thus the integer solution to (3) is
$$
m^{\prime}=406-13 k, n^{\prime}=-203+7 k, k \in \mathbf{Z}
$$

Let $m^{\prime} \geqslant 0, n \geqslant 0$ and solve to get $29 \leqslant k \leqslant 31$.
Take $k=29,30,31$, to get three sets of non-negative integer solutions to (3) that satisfy the conditions:
$$
\left\{\begin{array} { l } 
{ m ^ { \prime } = 2 9 } \\
{ n ^ { \prime } = 0 }
\end{array} \quad \left\{\begin{array} { l } 
{ m ^ { \prime } = 1 6 } \\
{ n ^ { \prime } = 7 }
\end{array} \quad \left\{\begin{array}{l}
m^{\prime}=3 \\
n^{\prime}=14
\end{array}\right.\right.\right.
$$

Thus, we get three sets of positive integer solutions to (2) that satisfy the conditions:
$$
\left\{\begin{array} { l } 
{ m = 3 3 } \\
{ n = 3 }
\end{array} \quad \left\{\begin{array} { l } 
{ m = 2 0 } \\
{ n = 1 0 }
\end{array} \quad \left\{\begin{array}{l}
m=7 \\
n=17
\end{array} .\right.\right.\right.
$$
(1) When $m=33, n=3$, it is obvious that $x_{5}=x_{6}=x_{7}=13$ has only one possibility;

Also, 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 $\mathrm{C}_{32-1}^{4-1}=\mathrm{C}_{32}^{3}=4960$ sets of positive integer solutions.
It is known that there are 4960 sets of positive integer solutions to (1) that satisfy the conditions at this time.
(2) When $m=20, n=10$, let $x_{i}=7 y_{i}(i=1,2,3,4), x_{j}=7 y_{j}(j=5,6,7)$,

By $y_{1}+y_{2}+y_{3}+y_{4}=20$, there are $\mathrm{C}_{19}^{3}$ sets of positive integer solutions: and $y^{5}+y^{6}+y^{7}=10$, there are $\mathrm{C}_{9}^{2} \mathrm{C}_{9}^{1}$ sets of positive integer solutions.
It is known that there are $\mathrm{C}_{19}^{3} \mathrm{C}_{9}^{2}=34884$ sets of positive integer solutions to (1) that satisfy the conditions at this time.
(3) When $m=7, n=17$, let $x_{i}=7 y_{i}(i=1,2,3,4), x_{j}=7 y_{j}(j=5,6,7)$,

By $y^{1}+y^{2}+y^{3}+y^{4}=7$ and $y^{5}+y^{6}+y^{7}=17$, there are $\mathrm{C}_{6}^{3}$ and $\mathrm{C}_{16}^{2}$ sets of positive integer solutions, respectively. It is known that there are $\mathrm{C}_{16}^{2} \mathrm{C}_{6}^{3}=2400$ sets of positive integer solutions to (1) that satisfy the conditions at this time.
In summary, the number of sets of positive integer solutions to (1) that satisfy the conditions is:
$$
\mathrm{C}_{32}^{3}+\mathrm{C}_{19}^{3} \mathrm{C}_{9}^{2}+\mathrm{C}_{16}^{2} \mathrm{C}_{6}^{3}=4960+34884+2400=42244
$$