, 2015

Solution:
Answer: 24

Note that $n^{0} = \binom{n}{0}$, $n^{1} = \binom{n}{1}$, $n^{2} = 2\binom{n}{2} + \binom{n}{1}$, $n^{3} = 6\binom{n}{3} + 6\binom{n}{2} + \binom{n}{1}$ (generally see http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind). Thus the polynomial rewrites as
$$6A \binom{n}{3} + (6A + 2B) \binom{n}{2} + (A + B + 2C) \binom{n}{1} + D \binom{n}{0},$$
which by the classification of integer-valued polynomials is divisible by $N$ always if and only if $6A$, $6A + 2B$, $A + B + 2C$, $D$ are always divisible by $N$.

We can eliminate $B$ and (trivially) $D$ from the system: it's equivalent to the system $6A \equiv 0 \pmod{N}$, $4A - 4C \equiv 0 \pmod{N}$, $B \equiv -A - 2C \pmod{N}$, $D \equiv 0 \pmod{N}$. So we want $1^2$ times the number of $(A, C)$ with $A \equiv 0 \pmod{N/6}$, $C \equiv A \pmod{N/4}$. So there are $N/(N/6) = 6$ choices for $A$, and then given such a choice of $A$ there are $N/(N/4) = 4$ choices for $C$. So we have $6 \cdot 4 \cdot 1^2 = 24$ solutions total.