For the polynomial
$$
p(x)=985 x^{2021}+211 x^{2020}-211 \text {, }
$$

let its 2021 complex roots be $x_{1}, x_{2}, \cdots, x_{2021}$. Then
$$
\sum_{k=1}^{2021} \frac{1}{x_{k}^{2}+1}=
$$
$\qquad$
4. 2021 .
$$
\begin{array}{l}
\text { Let } p(x)=985\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{2001}\right) . \\
\text { Then } p^{\prime}(x)=985 \times 2021 x^{2000}+211 \times 2020 x^{2019}, \\
\frac{p^{\prime}(x)}{p(x)}=\sum_{k=1}^{2021} \frac{1}{x-x_{k}} . \\
\text { Also, } \frac{1}{x_{k}^{2}+1}=\frac{\frac{1}{-\mathrm{i}-x_{k}}-\frac{1}{\mathrm{i}-x_{k}}}{2 \mathrm{i}}, \text { thus, } \\
\sum_{k=1}^{2021} \frac{1}{x_{k}^{2}+1}=\frac{\frac{p^{\prime}(-\mathrm{i})}{p(-\mathrm{i})}-\frac{p^{\prime}(\mathrm{i})}{p(\mathrm{i})}}{2 \mathrm{i}} \\
\Rightarrow \frac{1}{2 \mathrm{i}}\left(\frac{985 \times 2021(-\mathrm{i})^{2000}+211 \times 2000(-\mathrm{i})^{2019}}{985(-\mathrm{i})^{2021}+211(-\mathrm{i})^{2000}-211}-\right. \\
\left.\frac{985 \times 2021 \mathrm{i}^{2020}+211 \times 2020 \mathrm{i}^{2019}}{985 \mathrm{i}^{2021}+211 \mathrm{i}^{2020}-211}\right) \\
=2021 .
\end{array}
$$