The given function \(f: \mathbf{R} \rightarrow \mathbf{R}\) is define as
\(\begin{aligned}
& f(x)=\frac{2020^x}{2020^x+\sqrt{2020}}, \forall x \in \mathbf{R} \\
& \therefore f(1-x)+f(x)=\frac{\sqrt{2020}}{\sqrt{2020}+\sqrt{2020^x}} +\frac{2020^x}{2020^x+\sqrt{2020}}=1 \\
& \therefore 2 \sum_{r=1}^{4039} f\left(\frac{r}{4040}\right)=2 \\
& {\left[\left\{f\left(\frac{1}{4040}\right)+f\left(\frac{4039}{4040}\right)\right\}+\left\{f\left(\frac{2}{4040}\right)+f\left(\frac{4038}{4040}\right)\right\}\right]}\left.+\ldots .+\left\{f\left(\frac{2019}{4040}\right)+f\left(\frac{2021}{4040}\right)\right\}+f\left(\frac{2020}{4040}\right)\right] \\
& =2\left[2019+f\left(\frac{1}{2}\right)\right]=2\left[2019+\frac{1}{2}\right]\left\{\because f\left(\frac{1}{2}\right)=\frac{1}{2}\right\} \\
& =4039
\end{aligned}\)
Hence, option (b) is correct.