Note that $2015=5 \times 13 \times 31$ and that $N=2^{30} \cdot 3^{12} \cdot 5^{4}$ has exactly 2015 positive factors. We claim this is the smallest such integer. Note that $N<2^{66}$. If $n$ has 3 distinct prime factors, it must be of the form $p^{30} q^{12} r^{4}$ for some primes $p, q, r$, so $n \geq 2^{30} \cdot 3^{12} \cdot 5^{4}$. If $n$ has 2 distinct prime factors, it must be of the form $p^{e} q^{f}>2^{e+f}$ where $(e+1)(f+1)=2015$. It is easy to see that this means $e+f>66$ so $n>2^{66}>N$. If $n$ has only 1 prime factor, we have $n \geq 2^{2014}>N$. So $N$ is the smallest such integer, and the sum of its prime factors is $2 \cdot 30+3 \cdot 12+5 \cdot 4=116$.