Three. (50 points)
Before the World Cup, the coach of country $F$ plans to evaluate 7 players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in 3 training matches (each 90 minutes long). Assume that at any moment during the matches, only one of these players is on the field, and the total playing time (in minutes) for $A_{1}, A_{2}, A_{3}, A_{4}$ is divisible by 7, while the total playing time (in minutes) for $A_{5}, A_{6}, A_{7}$ 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?
Three, (50 points)
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 we have
$$
\sum_{i=1}^{4} x_{i}=7 m, \sum_{i=5}^{?} x_{i}=13 n, m, n \in N .
$$

Thus, $m, n$ are positive integer solutions to the indeterminate equation $7 m+13 n=270$
under the conditions $m \geqslant 4$ and $n \geq 3$.
Since $7(m-4)+13(n-3)=203$.
Let $m^{\prime}=m-4, n^{\prime}=n-3$, we have
$$
7 m^{\prime}+13 n^{\prime}=203 \text {. }
$$

Therefore, finding (2) satisfying the conditions $m \geqslant 4, n \geq 3$ is equivalent to finding (3) with non-negative integer solutions.
It is easy to observe that
$$
\begin{array}{c}
7 \cdot 2+13 \cdot(-1)=1, \\
7 \cdot 406+13 \cdot(-203)=203
\end{array}
$$

Thus, $m_{0}=406 n_{0}=203$ is a particular integer solution to (3), and the general integer solution to (3) is
$$
m^{\prime}=406-13 k, \quad n^{\prime}=-203+7 k \quad k \in \mathbf{Z} .
$$

Let $m^{\prime} \geqslant 0, n^{\prime} \geqslant 0$, solving gives $29 \leqslant k \leqslant 31$.
Taking $k=29,30,31$, we get 3 sets of non-negative integer solutions to (3):
$$
\left\{\begin{array} { l } 
{ m ^ { \prime } = 2 9 } \\
{ n ^ { \prime } = 0 }
\end{array} \quad \left\{\begin{array} { l } 
{ m ^ { \prime } = 1 6 } \\
{ n ^ { \prime } = 7 }
\end{array} \quad \left\{\begin{array}{l}
m^{\prime}=3 \\
n^{\prime}=14
\end{array}\right.\right.\right.
$$

From this, we get 3 sets of positive integer solutions to (2):
$$
\left\{\begin{array} { l } 
{ m = 3 3 } \\
{ n = 3 }
\end{array} \left\{\begin{array} { l } 
{ m = 2 0 } \\
{ n = 1 0 }
\end{array} \quad \left\{\begin{array}{l}
m=7 \\
n=17
\end{array}\right.\right.\right.
$$
(1) When $m=33, n=3$, it is clear that $x_{4}=x_{6}=x_{7}=13$ has only one possibility;
For $x_{2}=7 y_{i}(i=1,2,3,4)$, the indeterminate equation $y_{1}+y_{2}+y_{3}+y_{4}=33$ has $C_{33-1}^{4-1}=C_{32}^{3}=4960$ sets of positive integer solutions.
Thus, in this case, (1) has $C_{32}^{3}=4960$ sets of positive integer solutions.
(2) When $m=20, n=10$, let $x_{t}=7 y_{i}\{i=1,2,3,4), x_{j}=13 y_{j}(j=5,6, 7)$,
the equation $y_{1}+y_{2}+y_{3}+y_{4}=20$ has $C_{19}^{3}$ sets of positive integer solutions;
and $y_{5}+y_{6}+y_{7}=10$ has $C_{9}^{2}$ sets of positive integer solutions.
Thus, in this case, (1) has $C_{19}^{3} \cdot C_{9}^{2}=34884$ sets of positive integer solutions.
(3) When $m=7, n=17$, still let $x_{i}=7 y_{i}(i=1,2,3,4), x_{j}=13 y_{i}(j=5,6, 7)$, the equations $y_{1}+y_{2}+y_{3}+y_{4}=7$ and $y_{5}+y_{6}+y_{7}=17$ have $C_{6}^{3}$ and $C_{16}^{2}$ sets of positive integer solutions, respectively.
Thus, in this case, (1) has $C_{6}^{3} \cdot C_{16}^{2}=2400$ sets of positive integer solutions.
In summary, the number of sets of positive integer solutions to (1) is
$$
C_{32}^{3}+C_{19}^{3} \cdot C_{9}^{2}+C_{6}^{3} \cdot C_{16}^{2}=4960+34884+2400=42244 .
$$