6. D.

Given that $5 n^{2}+3 n-5$ is a multiple of 15
$$
\Rightarrow 5\left|\left(5 n^{2}+3 n-5\right) \Rightarrow 5\right| 3 n \Rightarrow 5 \mid n \text {. }
$$

Let $n=5 m\left(m \in \mathbf{N}_{+}\right)$. Then
$$
5 n^{2}+3 n-5=120 m^{2}+15 m+5\left(m^{2}-1\right) \text {. }
$$

Since $5 n^{2}+3 n-5$ is a multiple of 15, it follows that $m^{2}-1$ is a multiple of 3.
Thus, $m=3 k+1$ or $m=3 k+2(k \in \mathbf{N})$.
Therefore, $n=15 k+5$ or $15 k+10(k \in \mathbf{N})$.
Hence, the sum of all positive integers $n$ that satisfy the condition is
$$
\begin{array}{c}
(5+20+35+50+65+80+95)+ \\
(10+25+40+55+70+85)=635 .
\end{array}
$$
6. Given that $n$ is a positive integer less than 100, and makes $5 n^{2}+3 n-5$ a multiple of 15. Then the sum of all positive integers $n$ that satisfy this condition is ( ).
(A) 285
(B) 350
(C) 540
(D) 635