3. Before the World Cup, the coach of country $F$ wants to evaluate the seven players $A_{1}, A_{2}, \cdots, A_{7}$, and plans to have them all play in three training matches (each match lasting 90 minutes). 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 7, while the total playing time (in minutes) of $A_{5}, A_{6}, A_{7}$ 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?
Let the playing time of the $i$-th player be $x_{i}(i=1,2, \cdots, 7)$ minutes, then $x_{1}+x_{2}+\cdots+x_{7}=270$, where $7 \mid x_{i}(1 \leqslant i \leqslant 4)$ and $13 \mid x_{j}(5 \leqslant j \leqslant 7)$. Let $\sum_{i=1}^{4} x_{i}=7 m, \sum_{j=5}^{7} x_{j}=13 n$, thus $m \geqslant 4, n \geqslant 3,7 m+13 n=270 \Rightarrow 7(m-4)+13(n-3)=203$. Denote $m^{\prime}=m-4, n^{\prime}=n-3$, then the particular solution of $7 m^{\prime}+13 n^{\prime}=203$ is $(406,-203)$, and the general solution is $(406-13 t,-203+7 t)$. Considering $\left\{\begin{array}{l}406-13 t \geqslant 0, \\ -203+7 t \geqslant 0\end{array} \Rightarrow 29 \leqslant t \leqslant 31\right.$. Therefore, the 3 solutions satisfying $7 m^{\prime}+13 n^{\prime}=203$ are $(29,0),(16,7),(3,14)$, thus the 3 solutions satisfying $7 m+13 n=270$ are $(33,3),(20,10),(7,17)$.
(1) When $m=33, n=3$, $x_{5}=x_{6}=x_{7}=13$, let $x_{i}=7 y_{i}(i=1,2,3,4)$, then the positive integer solutions of $y_{1}+y_{2}+y_{3}+y_{4}=33$ are $C_{32}^{3}=4960$ groups, at this time the positive integer solutions satisfying the condition are 4960 groups;
(2) When $m=20, n=10$, let $x_{i}=7 y_{i}(i=1,2,3,4), x_{j}=13 y_{j}(j=5,6,7)$, then the positive integer solutions of $y_{1}+y_{2}+y_{3}+y_{4}=20$ are $C_{19}^{3}$ groups, the positive integer solutions of $y_{5}+y_{6}+y_{7}=10$ are $C_{9}^{2}$ groups, at this time the positive integer solutions satisfying the condition are $C_{19}^{3} \cdot C_{9}^{2}=34884$ groups;
(3) When $m=7, n=17$, then the positive integer solutions of $y_{1}+y_{2}+y_{3}+y_{4}=7$ are $C_{6}^{3}$ groups, the positive integer solutions of $y_{5}+y_{6}+y_{7}=$ 17 are $C_{16}^{2}$ groups, at this time the positive integer solutions satisfying the condition are $C_{6}^{3} \cdot C_{16}^{2}=2400$ groups.
In summary, the positive integer solutions satisfying the condition are $4960+34884+2400=42244$ groups.