Suppose that $x_i$ ($i = 1, 2, \dots, 7$) minutes is the time for $i$-th player on the field. Now, the problem is to find the number of solution groups of positive integers for the following equation:
$$x_1 + x_2 + \cdots + x_7 = 270$$
when the conditions $7 \mid x_i$ ($i = 1, 2, 3, 4$) and $13 \mid x_j$ ($j = 5, 6, 7$) are satisfied.

Suppose $x_1 + x_2 + x_3 + x_4 = 7m$ and $x_5 + x_6 + x_7 = 13n$. Then
$$7m + 13n = 270,$$
and $m, n \in \mathbb{N}_+, m \ge 4$ and $n \ge 3$.

When $(m, n) = (33, 3)$, $x_5 = x_6 = x_7 = 13$. Let $x_i = 7y_i$ ($i = 1, 2, 3, 4$), then
$$y_1 + y_2 + y_3 + y_4 = 33.$$
We get $C_{33-1}^4 = C_{32}^3 = 4960$ solution groups of positive integers $(y_1, y_2, y_3, y_4)$ and in this case, we have $4960$ solution groups of positive integers satisfying the conditions.

When $(m, n) = (20, 10)$, let $x_i = 7y_i$ ($i = 1, 2, 3, 4$) and $x_j = 13y_j$ ($j = 5, 6, 7$). Hence
$$y_1 + y_2 + y_3 + y_4 = 20 \text{ and } y_5 + y_6 + y_7 = 10.$$
In this case, we have $C_{19}^3 \times C_9^3 = 34884$ solution groups of positive integers satisfying the conditions.

When $(m, n) = (7, 17)$, set $x_i = 7y_i$ ($i = 1, 2, 3, 4$) and $x_j = 13y_j$ ($j = 5, 6, 7$). Hence
$$y_1 + y_2 + y_3 + y_4 = 7 \text{ and } y_5 + y_6 + y_7 = 17.$$
In this case, we have $C_6^3 \times C_{16}^3 = 2400$ solution groups of positive integers satisfying the conditions.

Consequently, for $(1)$, there are
$$4960 + 34884 + 2400 = 42244$$
solution groups of positive integers satisfying the conditions.