　${\dfrac{1}{s_i}+\dfrac{1}{s_{2021-i}}=1}$ を変形して $(s_i-1)(s_{2021-i}-1)=1$ を得るから,

$$
\begin{aligned}
\sum_{i=1}^{2021}\dfrac{1}{{s_i}^3-3{s_i}^2+3{s_i}}&=\sum_{i=1}^{2021}\dfrac{1}{(s_i-1)^3+1}\\\\
&=1+\sum_{i=1}^{1010}\left(\frac{1}{(s_i-1)^3+1}+\frac{1}{(s_{2021-i}-1)^3+1}\right)\\\\
&=1+\sum_{i=1}^{1010}\left(\frac{1}{(s_i-1)^3+1}+\cfrac{1}{\cfrac{1}{(s_i-1)^3}+1}\right)\\\\
&=1+\sum_{i=1}^{1010}1 = \textbf{1011}
\end{aligned}
$$
公式解説
https://onlinemathcontest.com/contests/omckoukou1/editorial/2194
　$s_i=\dfrac{2021}{i}$ について, 以下の総和を求めてください：
$$\displaystyle\sum_{i=1}^{2021}\frac{1}{{s_i}^3-3{s_i}^2+3{s_i}}$$