Solution:

The answer is $\mathbf{( C )}$. If $m$ is a multiple of a certain prime $p$, then it must be divisible by the maximum power of $p$ dividing $25!$ (where by $25!$ we mean the product of the integers from 1 to 25) so that $n=\frac{25!}{m}$ has no factors of $p$ (otherwise the fraction $\frac{m}{n}$ would not be reduced to lowest terms). We must therefore count the divisors $m$ of $25!$ such that

a. $m$ contains in its factorization some of the prime factors of $25!$, each raised to the same power to which it appears in the factorization of $25!$ and

b. $m<\frac{25!}{m}$.

Pairing each divisor $d$ having property (a) with the divisor $\frac{25!}{d}$, since between the two only the smaller one will have property (b) (given our requirements on the prime factors of $d$, we cannot have $d=\frac{25!}{d}$), we obtain that the divisors to be counted will be half of those for which only property (a) is required.

In the factorization of $25!$ there appear 9 distinct primes: $2,3,5,7,11,13,17,19,23$. The divisors with property (a) are therefore $2^{9}$, and among these $2^{8}$ have property (b).