Three, (50 points) Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in minutes) for $A_{1}, A_{2}, A_{3}, A_{4}$ can all be divided by 7, while the total playing time (in minutes) for $A_{5}, A_{6}, A_{7}$ can all be divided 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, 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 $71 x_{i}(1 \leqslant i \leqslant 4)$ and $131 x_{j}(5 \leqslant j \leqslant 7)$.

If $\left(x_{1}, x_{2}, \cdots, x_{7}\right)$ is a set of positive integer solutions satisfying condition (1), then it should have
$$
\sum_{i=1}^{4} x_{i}=7 m, \sum_{i=5}^{7} x_{i}=13 n, m, n \in \mathbf{N} .
$$

Thus, $m$ and $n$ are a set of positive integer solutions to the indeterminate equation
$$
7 m+13 n=270
$$

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
$$
7 m^{\prime}+13 n^{\prime}=203 \text {. }
$$

Therefore, finding (2) satisfying the conditions $m \geqslant 4, n \geqslant 3$ is equivalent to finding the non-negative integer solutions to (3).
It is easy to observe that $7 \times 2+13 \times(-1)=1$. Thus, we have
$$
7 \times 406+13 \times(-203)=203 \text {, }
$$

which means $m_{0}=406, n_{0}=-203$ is a particular integer solution to (3).
Thus, the general integer solution to (3) is
$$
\begin{array}{l}
m^{\prime}=406-13 k, n^{\prime}=-203+7 k, k \in \mathbf{Z} . \\
\text { Let } m^{\prime} \geqslant 0, n^{\prime} \geqslant 0, \text { solving gives } 29 \leqslant k \leqslant 31 .
\end{array}
$$

Taking $k=29,30,31$, we 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 clear that $x_{5}=x_{6}=x_{7}=13$ has only one possibility;

Let $x_{i}=7 y_{i}(i=1,2,3,4)$, then, by the indeterminate equation $y_{i}+y_{2}+y_{3}+y_{4}=33$, there are $\mathrm{C}_{33-1}^{4-1}=\mathrm{C}_{32}^{3}=4960$ sets of positive integer solutions. Thus, in this case, (1) has $\mathrm{C}_{32}^{3}=4960$ sets of positive integer solutions,
2) When $m=20, n=10$, let $x_{1}=7 y_{1}(i=1,2,3,4)$, $x_{j}=13 y_{j}(j=5,6,7)$. By $y_{1}+y_{2}+y_{3}+y_{4}=20$, there are $\mathrm{C}_{19}^{3}$ sets of positive integer solutions, and by $y_{5}+y_{6}+y_{7}=10$, there are $\mathrm{C}_{9}^{2}$ sets of positive integer solutions. Thus, in this case, (1) has $\mathrm{C}_{19}^{3} \cdot \mathrm{C}_{9}^{2}=34884$ sets of positive integer solutions.
3) When $m=7, n=17$, still let $x_{i}=7 y_{i}(i=1,2,3,4)$, $x_{j}=13 y_{1}(j=5,6,7)$. 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. Thus, in this case, (1) has $\mathrm{C}_{6}^{3} \cdot \mathrm{C}_{10}^{2}=2400$ sets of positive integer solutions.
In summary, the number of sets of positive integer solutions to (1) that satisfy the conditions is
$$
C_{32}^{3}+C_{19}^{3} \cdot C_{9}^{2}+C_{6}^{3} \cdot C_{16}^{2}=42244 .
$$