We can compute that $1+2+\cdots+n=\frac{n(n+1)}{2}$ and $(n+1)^{2}+(n+2)^{2}+\cdots+(2 n)^{2}=\frac{2 n(2 n+1)(4 n+1)}{6}-\frac{n(n+1)(2 n+1)}{6}=\frac{n(2 n+1)(7 n+1)}{6}$, so we need $\frac{15(2 n+1)(7 n+1)}{3(n+1)}=\frac{5(2 n+1)(7 n+1)}{n+1}$ to be an integer. The remainder when $(2 n+1)(7 n+1)$ is divided by $(n+1)$ is 6, so after long division we need $\frac{30}{n+1}$ to be an integer. The solutions are one less than a divisor of 30 so the answer is $$1+2+4+5+9+14+29=64$$