Solution:

Since $2004 = 2^2 \cdot 3 \cdot 167,$ we have $2004^{2004} = 2^{4008} \cdot 3^{2004} \cdot 167^{2004},$ so its divisors are $N = 2^i 3^j 167^k$ with $i \le 4008$ and $j, k \le 2004.$ Such an $N$ has $(i+1)(j+1)(k+1)$ divisors, so we need $(i+1)(j+1)(k+1) = 2004.$

Every ordered triple of positive integers with product $2004$ yields admissible exponents, since each factor is at most $2004.$ Counting prime by prime: the exponent $2$ of the prime $2$ is split among the three factors in $\binom{2+2}{2} = 6$ ways by stars and bars, and each of the primes $3$ and $167$ goes to one of the $3$ factors.

The count is $6 \cdot 3 \cdot 3 = 54.$