Example 3.3.5 Let $T$ be the set of all positive divisors of $2004^{100}$, and $S$ be a subset of $T$, where no element in $S$ is an integer multiple of any other element in $S$. Find the maximum value of $|S|$.
Solution: $2004=2^{2} \times 3 \times 167$, so
$T=\left\{2^{a} 3^{b} 167^{c} ; 0 \leqslant a \leqslant 200,0 \leqslant b \leqslant 100,0 \leqslant c \leqslant 100, \forall a, b, c \in \mathbf{N}\right\}$;
Since $\{(b, c) ; 0 \leqslant b, c \leqslant 100, b, c \in \mathbf{N}\}$ has $101^{2}$ elements, if $S \subseteq T$ and $|S|>101^{2}$, then by the pigeonhole principle, $S$ must have two numbers corresponding to the same integer pair $b, c$, let these two numbers be
$$
s_{1}=2^{a_{1}} 3^{b} 167^{c}, s_{2}=2^{a_{2}} 3^{b} 167^{c}.
$$

If $a_{1}>a_{2}$, then $s_{2} \mid s_{1}$; if $a_{1}<a_{2}$, then $s_{1} \mid s_{2}$.
Therefore, such an $S$ does not meet the problem's requirements, and for an $S$ that satisfies the requirements, $|S| \leqslant 101^{2}$.
Now consider $S_{0}=\left\{2^{200-b-c} 3^{b} 167^{c} ; 0 \leqslant b, c \leqslant 100 ; b, c \in \mathbf{N}\right\}$, it is easy to see that $S_{0} \subseteq T$, and $\left|S_{0}\right|=101^{2}$. We now prove that $S_{0}$ meets the problem's requirements.
If not, there are two different elements in $S_{0}$
$$
s_{1}=2^{200-b_{1}-c_{1}} 3^{b_{1}} 167^{c_{1}}, s_{2}=2^{200-b_{2}-c_{2}} 3^{b_{2}} 167^{c_{2}}
$$

where one number is an integer multiple of the other, without loss of generality, assume $s_{2} \mid s_{1}$, then we have
$$
200-b_{1}-c_{1} \geqslant 200-b_{2}-c_{2}, b_{1} \geqslant b_{2}, c_{1} \geqslant c_{2}.
$$

From the first inequality, we have $b_{1}+c_{1} \leqslant b_{2}+c_{2}$, from the second and third inequalities, we have $b_{1}+c_{1} \geqslant b_{2}+c_{2}$, so $b_{1}+c_{1}=b_{2}+c_{2}$, and combining with $b_{1} \geqslant b_{2}, c_{1} \geqslant c_{2}$, we get $b_{1}=b_{2}, c_{1}=c_{2}, s_{1}=s_{2}$, which is a contradiction.
Therefore, for an $S$ that satisfies the conditions, the maximum value of $|S|$ is $101^{2}=10201$.