One, (40 points) Given $a_{1}, a_{2}, \cdots, a_{n}$ is an arithmetic sequence, satisfying
$$
\sum_{i=1}^{n}\left|a_{i}\right|=\sum_{i=1}^{n}\left|a_{i}-1\right|=\sum_{i=1}^{n}\left|a_{i}+1\right|=2021 .
$$

Find the maximum value of $n$.
Let's assume
$$
a_{1}<a_{2}<\cdots<a_{k} \leqslant 0<a_{k+1}<\cdots<a_{n} .
$$

Then $k=\left|a_{1}-1\right|-\left|a_{1}\right|+\left|a_{2}-1\right|-$
$$
\begin{aligned}
& \left|a_{2}\right|+\cdots+\left|a_{k}-1\right|-\left|a_{k}\right| \\
= & \left|a_{k+1}\right|-\left|a_{k+1}-1\right|+\cdots+\left|a_{n}\right|-\left|a_{n}-1\right| \\
\leqslant & n-k \\
\Rightarrow & 2 k \leqslant n,
\end{aligned}
$$

with equality holding if and only if $a_{k+1} \geqslant 1$.
$$
\begin{aligned}
\text { Also, } & n-k \\
= & \left|a_{k+1}+1\right|-\left|a_{k+1}\right|+\left|a_{k+2}+1\right|- \\
& \left|a_{k+2}\right|+\cdots+\left|a_{n}+1\right|-\left|a_{n}\right| \\
= & \left|a_{1}\right|-\left|a_{1}+1\right|+\cdots+\left|a_{k}\right|-\left|a_{k}+1\right| \\
\leqslant & k,
\end{aligned}
$$

Thus, $2 k \geqslant n$, with equality holding if and only if $a_{k} \leqslant-1$.
In summary, $n=2 k$, and $a_{k} \leqslant-1, a_{k+1} \geqslant 1$.
Let the common difference of the sequence be $d$. Then
$$
\begin{array}{l}
d=a_{k+1}-a_{k} \geqslant 2 . \\
\text { Hence } 2021=\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{2 k}\right| \\
=a_{k+1}-a_{k}+a_{k+2}-a_{k-1}+\cdots+a_{2 k}-a_{1} \\
=d+3 d+\cdots+(2 k-1) d=k^{2} d \geqslant 2 k^{2} .
\end{array}
$$

Therefore, $k^{2} \leqslant 1010$.
Thus, the maximum value of $k$ is 31.
When $k=31$, take $d=\frac{2021}{961}, a_{31}=-1$. Therefore, the maximum value of $n$ is 62.