Solution:

We first note that all the prime factors of $n$ must be $1$ modulo $3$ (and thus $1$ modulo $6$). The smallest primes with this property are $7, 13, 19, \ldots$. Since $7^{4} = 2401 > 1000$, the number can have at most $3$ prime factors (including repeats). Since $7 \cdot 13 \cdot 19 = 1729 > 1000$, the most factors $N$ can have is $6$.

Consider the number $7^{2} \cdot 19 = 931$, which has $6$ factors. For this choice of $N$, $N \cdot d(N) = 5586$.

For another $N$ to do better, it must have at least $6$ factors, for otherwise, $N \cdot d(N) < 1000 \cdot 5 = 5000$. It is easy to verify that $7^{2} \cdot 19$ is the greatest number with $6$ prime factors satisfying our conditions, so the answer must be $5586$.