Four, (50 points) Let $T$ be the set of all positive divisors of $2020^{100}$, and set $S$ satisfies:
(1) $S$ is a subset of $T$;
(2) No element in $S$ is a multiple of another element in $S$.
Find the maximum number of elements in $S$.
Because $2020=2^{2} \times 5 \times 101$, we have:
$$
\begin{aligned}
T= & \left\{2^{a} 5^{b} 101^{c} \mid 0 \leqslant a \leqslant 200,0 \leqslant b, c \leqslant 100,\right. \\
& a, b, c \in \mathbf{N}\} .
\end{aligned}
$$

Let $S=\left\{2^{200-b-c} 5^{b} 101^{c} \mid 10 \leqslant b, c \leqslant 100\right.$, $b, c \in \mathbf{N}\}$.

For any $0 \leqslant b, c \leqslant 100$, we have
$$
200-b-c \leqslant 200 \text {, }
$$

Thus, $S$ is a subset of $T$, and $S$ contains
$$
101 \times 101=101^{2}
$$

distinct elements.
The following proves that no element in $S$ is a multiple of another, and there is no other suitable set with more elements than $S$.

Assume that $2^{200-b-c} 5^{b} 101^{c}$ in $S$ is a multiple of $2^{200-j-k} 5^{j} 101^{k}$. Then
$$
200-b-c \geqslant 200-j-k, b \geqslant j, c \geqslant k .
$$

From the first inequality, we get $b+c \leqslant j+k$,
from the last two inequalities, we get $b+c \geqslant j+k$.
Thus, $b=j, c=k$.
Therefore, no element in $S$ is a multiple of another.

Assume $U$ is a subset of $T$ and $U$ contains more than $101^{2}$ elements.

Since the pair $(b, c)$ has only $101^{2}$ possible values, $U$ must contain two elements
$u_{1}=2^{a_{1}} 5^{b_{1}} 101^{c_{1}}$ and $u_{2}=2^{a_{2}} 5^{b_{2}} 101^{c_{2}}$, with $b_{1}=b_{2}, c_{1}=c_{2}$ but $a_{1} \neq a_{2}$.

If $a_{1}>a_{2}$, then $u_{1}$ is a multiple of $u_{2}$; if $a_{1}<a_{2}$, then $u_{2}$ is a multiple of $u_{1}$. This indicates that $U$ does not satisfy the given condition.
Therefore, the maximum number of elements in $T$ is $101^{2}=10201$.