How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers?
The [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization) of 2004 is $2^2\cdot 3\cdot 167$.  Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$.
We can [ count the number of divisors](https://artofproblemsolving.com/wiki/index.php/Divisor_function) of a number by multiplying together one more than each of the [ exponents](https://artofproblemsolving.com/wiki/index.php/Exponentiation) of the prime factors in its prime factorization.  For example, the number of divisors of $2004=2^2\cdot 3^1\cdot 167^1$ is $(2+1)(1+1)(1+1)=12$.  
A positive integer divisor of $2004^{2004}$ will be of the form $2^a\cdot 3^b\cdot 167^c$.  Thus we need to find how many $(a,b,c)$ satisfy 

$(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167.$
We can think of this as [partitioning](https://artofproblemsolving.com/wiki/index.php/Partition) the exponents to $a+1,$ $b+1,$ and  $c+1$.  So let's partition the 2's first.  There are two 2's so this is equivalent to partitioning two items in three containers.  We can do this in ${4 \choose 2} = 6$ ways.  We can partition the 3 in three ways and likewise we can partition the 167 in three ways.  So we have $6\cdot 3\cdot 3 = \boxed{54}$ as our answer.