, 2015

Solution:
Answer: $116$

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 $2 \cdot 30 + 3 \cdot 12 + 5 \cdot 4 = 116$.