Solution:

Since $17017 = 7 \cdot 11 \cdot 13 \cdot 17,$ the divisors of $17017^{17}$ are $7^a 11^b 13^c 17^d$ with each exponent between $0$ and $17.$ Modulo $12$ we have $13 \equiv 1,$ and $7^2 \equiv 11^2 \equiv 17^2 \equiv 1$ (as $17 \equiv 5$), so the residue of a divisor is $7^{\alpha} \, 11^{\beta} \, 5^{\delta} \pmod{12},$ where $\alpha, \beta, \delta$ are the parities of $a, b, d.$

The four possible values $1, 5, 7, 11$ multiply like the group $\{1, 5, 7, 11\}$ mod $12,$ in which $7 \cdot 11 \equiv 5.$ Checking the eight parity patterns, the residue is $5$ exactly when $(\alpha, \beta, \delta) = (0, 0, 1)$ or $(1, 1, 0).$ Each parity condition is satisfied by $9$ of the $18$ choices of that exponent, while $c$ is free with $18$ choices.

Therefore $N = 2 \cdot 9 \cdot 9 \cdot 9 \cdot 18 = 26244,$ and the remainder mod $1000$ is $244.$