Solution. The prime factorization of 2004 is: $2004=2^{2} \cdot 3 \cdot 167$, so if $2004^{n}$ is a divisor of 2004!, then $2^{2 n} \cdot 3^{n} \cdot 167^{n}$ must also be a divisor of 2004!. The critical point is $167^{n}$, because 167 is the largest prime factor, and thus it appears the least number of times in the prime factorization of 2004!. Since 2004 = 12$\cdot$167, there are 12 numbers between 1 and 2004 that are divisible by 167, and these numbers are only divisible by the first power of 167. Therefore, in the prime factorization of 2004!, $167^{12}$ appears, so the value of $n$ cannot be greater than 12. (It is easy to see that 2 and 3 appear with exponents much larger than 24 and 12, respectively, in the prime factorization of 2004!, so 12 is also an acceptable exponent.) Therefore, there are 12 positive integers that satisfy the conditions of the problem.
How many positive integers can be written in place of $n$, if 2004n is a divisor of 2004! ? [2004! denotes the product of integers from 1 to 2004.]