10. Let $f(x)=\frac{1}{x^{3}-x}$, find the smallest positive integer $n$ that satisfies the inequality $f(2)+f(3)+\cdots+f(n)>\frac{499}{2020}$.
$$
\begin{array}{l}
f(n)=\frac{1}{n^{3}-n}=\frac{1}{n(n-1)(n+1)}=\frac{1}{2}\left[\frac{1}{(n-1) n}-\frac{1}{n(n+1)}\right], \\
\text { Thus } f(2)+f(3)+\cdots+f(n) \\
=\frac{1}{2}\left[\frac{1}{1 \times 2}-\frac{1}{2 \times 3}+\frac{1}{2 \times 3}-\frac{1}{3 \times 4}+\cdots+\frac{1}{(n-1) n}-\frac{1}{n(n+1)}\right] \\
=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{n(n+1)}\right]>\frac{499}{2020} \Rightarrow \frac{1}{2}-\frac{1}{n(n+1)}>\frac{499}{1010} \\
\Rightarrow \frac{1}{n(n+1)}<\frac{1}{1010} .
\end{array}
$$

Solving the equation $3 n^{2}+3 n-505=0$ yields the larger root $x_{2}=\frac{-3+\sqrt{9+12 \times 505}}{6}$. For $n=12, 3 \times 12^{2}+3 \times 12-505=-37<0$, so the smallest positive integer $n$ that satisfies the inequality is $n=13$.