Example 3 (36th Canadian Mathematical Olympiad) Let $T$ be the set of all positive divisors of $20044^{100}$, and let $S$ be a subset of $T$ such that no element of $S$ is an integer multiple of any other element of $S$. Find the maximum possible number of elements in $S$.

The set $T$ consists of all positive divisors of $20044^{100}$. A subset $S$ of $T$ is such that no element of $S$ is an integer multiple of any other element of $S$. Determine the maximum possible number of elements in $S$.
Given that $2004=2^{2} \times 3 \times 167$, we have:
$$
T=\left\{2^{a} \times 3^{b} \times 167^{c} \mid 0 \leqslant a \leqslant 200, 0 \leqslant b, c \leqslant 100, \forall a, b, c \in \mathbf{N}\right\}.
$$

Let $S=\left\{2^{p} \times 3^{q} \times 167^{r}\right\}$, where $p$ is some natural number not exceeding 200, and $q, r$ are some natural numbers not exceeding 100.
Construct a subset $S$ of $T$ that satisfies the given conditions:
When $q, r$ are determined, the value of $p$ is uniquely determined. Let $p=200-q-r$. When $q, r$ take all values from 0 to 100, i.e., $0 \leqslant q, r \leqslant 100$, $p$ satisfies $0 \leqslant p \leqslant 200$.
Thus, take $S=\left\{2^{200-b-c} \times 3^{b} \times 167^{c} \mid 0 \leqslant b, c \leqslant 100, \forall b, c \in \mathbf{N}\right\}$.
Since $b, c$ each have 101 possible values, we have $|S|=101^{2}$.
Next, we prove that no element in $S$ is a multiple of another, and that $T$ has no larger subset satisfying the given conditions.
(1) Suppose two elements in $S$ are $2^{200-b-c} \times 3^{b} \times 167^{c}$ and $2^{200-s-t} \times 3^{s} \times 167^{t}$, where $2^{200-b-c} \times 3^{b} \times 167^{c}$ is an integer multiple of $2^{200-s-t} \times 3^{s} \times 167^{t}$. Then,
$200-b-c \geqslant 200-s-t, b \geqslant s, c \geqslant t$.
From the first inequality, we get $b+c \leqslant s+t$;
From the second and third inequalities, we get $b+c \geqslant s+t$.
Thus, $b+c=s+t$.
Therefore, all the inequalities hold with equality, i.e., $b=s, c=t$. This is a contradiction.
Hence, no element in $S$ is a multiple of another.
(2) Suppose $U$ is a subset of $T$ with more than $101^{2}$ elements.
Since $0 \leqslant b, c \leqslant 100$, there are only $101^{2}$ different integer pairs $(b, c)$. Thus, by the pigeonhole principle, $U$ must contain at least two elements $u_{1}=2^{a_{1}} \times 3^{b_{1}} \times 167^{c_{1}}$ and $u_{2}=2^{a_{2}} \times 3^{b_{2}} \times 167^{c_{2}}$, such that $b_{1}=b_{2}$ and $c_{1}=c_{2}$, 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 $u_{1}$.
Thus, $U$ does not satisfy the given conditions.
Therefore, the maximum value is $101^{2}=10201$.