The residues of $n^3 \pmod{2015}$ cycle as $1, 3, 2, 4, 0, 1, 3, 2, 4, 0, ... \pmod{5},$ and the residues of $3^n \pmod{5}$ cycle as $3, 4, 2, 1, 3, 4, 2, 1, ... \pmod{5},$ so the residues of $n^3+3^n$ cyclic every $5 \cdot 4 = 20$ terms. In every group of $20,$ we see that there are $4$ ways to get a multiple of $5,$ and they are mainly $n^3 \equiv 1, 2, 3, 4$ and $3^n \equiv 4, 3, 2, 1$ respectively, and $\frac{2000}{20}=100,$ so there are $4 \cdot 100=400$ numbers that work from $1-2000.$ Now, we see the number of integers from $2001-2015,$ there are $3$ numbers that work, so the answer is $400+3=\boxed{403}.$