11. Let $n$ be a positive integer less than 100, and satisfies $\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n$ is an integer. Then the sum of all positive integers $n$ that meet the condition is $\qquad$
11. 635 .

Notice that,
$$
\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n=\frac{5 n^{2}+3 n-5}{15}
$$

is an integer, so, $15 \mid \left(5 n^{2}+3 n-5\right)$.
Thus, $5 \mid n$, and $3 \mid \left(n^{2}-1\right)$.
Therefore, $n=15 k+5$ or $15 k+10$.
Hence, the sum of all positive integers $n$ that satisfy the condition is
$$
\begin{array}{c}
\sum_{k=0}^{6}(15 k+5)+\sum_{k=0}^{5}(15 k+10) \\
=95+\sum_{k=0}^{5}(30 k+15) \\
=95+\frac{(15+165) \times 6}{2}=635 .
\end{array}
$$