Solution:

The answer is (D). We divide the 75 students of each of the two schools into 3 groups, according to the remainder of their number when divided by 3. In particular, in the first school, the students are divided into groups $A, B, C$ according to whether their number is of the form $3k, 3k+1$ or $3k+2$ for some integer $k$. Similarly in the second school the students are divided among $A', B', C'$.

Each of the groups $A, B, C, A', B', C'$ consists of 25 students, and group $A$ cannot play against group $A'$, just as $B$ cannot play against $B'$ and $C$ cannot play against $C'$.

Suppose that $m$ people from group $A$ play against someone in $B'$ and the remaining $25-m$ play against $C'$. Then $m$ people of $B$ play against the remaining $m$ in $C'$ and $25-m$ play against students of $A'$. Consequently there are $m$ people in $C$ who play against group $A'$ and the remaining $25-m$ play against group $B'$. Once $m$ is fixed, then, we choose $m$ people from $A, B, C$ in $\binom{25}{m}^3$ ways. Furthermore we choose in how many ways the people in $A'$ play against the $m$ chosen in $C$ and the $25-m$ in $B$ in $25!$ ways and similarly for $B', C'$. Varying $m$, then, the number of possible pairings is

$$n = (25!)^3 \sum_{m=0}^{25} \binom{25}{m}^3$$

We note that $\binom{25}{m}$ is divisible by 5 exactly if $m \neq 0$ and $m \neq 25$, hence

$$\binom{25}{m}^3 = 2 + 5h$$

is not a multiple of 5. The number $n$ therefore has as many factors of 5 as there are in $(25!)^3$, that is $6 \cdot 3 = 18$. We note that in $25!$ there are at least 6 factors of 2, for example because it includes $16 \cdot 8$, and therefore $n$ ends with exactly 18 zeros.