10. Let $S_{n}=1+2+\cdots+n$. Then among $S_{1}, S_{2}$, $\cdots, S_{2015}$, there are. $\qquad$ that are multiples of 2015.
10.8 .

Obviously, $S_{n}=\frac{1}{2} n(n+1)$.
For any positive divisor $d$ of 2015, it is easy to see that in the range $1 \sim 2015$, there is exactly one $n$ that satisfies $n$ being a multiple of $d$ and $n+1$ being a multiple of $\frac{2015}{d}$. Therefore, each divisor of 2015 will generate one $n$ such that $S_{n}$ is a multiple of 2015.

Since $2015=5 \times 13 \times 31$, 2015 has a total of eight divisors.
Thus, the desired result is 8.