Let $N=30^{2015}$.  Find the number of ordered 4-tuples of integers $(A,B,C,D)\in\{1,2,\ldots,N\}^4$ (not necessarily distinct) such that for every integer $n$, $An^3+Bn^2+2Cn+D$ is divisible by $N$.
1. Let \( N = 30^{2015} \). We need to find the number of ordered 4-tuples \((A, B, C, D) \in \{1, 2, \ldots, N\}^4\) such that for every integer \( n \), the polynomial \( An^3 + Bn^2 + 2Cn + D \) is divisible by \( N \).

2. First, consider the case when \( n = 0 \). Substituting \( n = 0 \) into the polynomial, we get:
   \[
   D \equiv 0 \pmod{N}
   \]
   Since \( D \) must be divisible by \( N \) and \( D \leq N \), the only possible value for \( D \) is \( D = N \).

3. Now, consider the polynomial \( An^3 + Bn^2 + 2Cn + D \) with \( D = N \):
   \[
   An^3 + Bn^2 + 2Cn + N
   \]
   For this polynomial to be divisible by \( N \) for all \( n \), the term \( An^3 + Bn^2 + 2Cn \) must also be divisible by \( N \). This implies:
   \[
   n(An^2 + Bn + 2C) \equiv 0 \pmod{N}
   \]
   Since \( n \) is not necessarily divisible by \( N \), the expression \( An^2 + Bn + 2C \) must be divisible by \( N \) for all \( n \).

4. Consider \( n = 0 \) again:
   \[
   2C \equiv 0 \pmod{N}
   \]
   This implies that \( C \) must be divisible by \( N/2 \). Since \( C \leq N \), the possible values for \( C \) are \( C = N \) or \( C = N/2 \).

5. Next, consider the polynomial \( An^2 + Bn + 2C \) for \( n = 1 \):
   \[
   A + B + 2C \equiv 0 \pmod{N}
   \]
   Since \( C \) can be either \( N \) or \( N/2 \), we need to check both cases:
   - If \( C = N \):
     \[
     A + B + 2N \equiv 0 \pmod{N} \implies A + B \equiv -2N \equiv 0 \pmod{N}
     \]
     This implies \( A \) and \( B \) must be multiples of \( N \). Since \( A, B \leq N \), the only possible values are \( A = N \) and \( B = N \).

   - If \( C = N/2 \):
     \[
     A + B + N \equiv 0 \pmod{N} \implies A + B \equiv -N \equiv 0 \pmod{N}
     \]
     This implies \( A \) and \( B \) must be multiples of \( N \). Since \( A, B \leq N \), the only possible values are \( A = N \) and \( B = N \).

6. Therefore, the only possible ordered 4-tuples \((A, B, C, D)\) are:
   \[
   (N, N, N, N) \quad \text{and} \quad (N, N, N/2, N)
   \]

The final answer is \(\boxed{2}\)