\begin{aligned}-、\left(40 \text { points) Given } a_{i}>0, i=1,2, \cdots, 2020,\right. \\ \text { and } \sum_{i=1}^{2020} a_{i}=2020 \text {. Find the maximum value of } \sum_{k=1}^{2020} a_{k}^{\frac{1}{k^{2}}} \text {. }\end{aligned}
$$
a_{k}+\underbrace{t+t+t}_{k^{2}-1 \uparrow} \geqslant k^{2} a_{k}^{\frac{1}{2^{2}}} t^{\frac{k^{2}-1}{k^{2}}}(k \geqslant 2) \text {. }
$$

Let $t=\frac{1}{k^{\frac{2 k^{2}}{k^{2}-1}}}(k \geqslant 2)$.
$$
\begin{array}{l}
\text { Then } a_{k}^{\frac{1}{k^{2}}} \leqslant a_{k}+\frac{k^{2}-1}{k^{\frac{22^{2}}{k^{2}-1}}} . \\
\text { Therefore } \sum_{k=1}^{2020} a_{k}^{\frac{1}{2}} \leqslant \sum_{k=1}^{2020} a_{k}+\sum_{k=1}^{2020} \frac{k^{2}-1}{k^{\frac{2 k 2}{k^{2}-1}}} \\
=2020+\sum_{k=1}^{2020} \frac{k^{2}-1}{k^{\frac{2 k^{2} 2}{k^{2}-1}}} .
\end{array}
$$

By $\frac{1}{k^{\frac{2 k^{2}}{22}-1}}0 .
$$

Let $a_{k}=\frac{1}{k^{\frac{2 k^{2}}{22}-1}}(k \geqslant 2)$,
$$
a_{1}=2020-a_{2}-a_{3}-\cdots-a_{2020},
$$

At this point, the equality in (1) holds.
Thus, the desired result is $2020+\sum_{k=1}^{2020} \frac{k^{2}-1}{k^{\frac{2 k 2}{k^{2}-1}}}$.