This question takes me back to 2004!

The 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 of a number by multiplying together one more than each of the exponents 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 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 $6\cdot 3\cdot 3 = 54$ as our answer.