Example 7 Let $T$ be the set of all divisors of $2004^{100}$, and $S$ be a subset of $T$ in which no number is a multiple of another. What is the maximum number of elements that $S$ can contain? ${ }^{[s]}$
$(2004$, Canadian Mathematical Olympiad)
Notice that $2004=2^{2} \times 3 \times 167$. Then
$$
T=\left\{2^{a} 3^{b} 167^{c} \mid 10 \leqslant a \leqslant 200, 0 \leqslant b, c \leqslant 100\right\} \text {. }
$$

Let $S=\left\{2^{200-b-c} 3^{b} 167^{c} \mid 0 \leqslant b, c \leqslant 100\right\}$.
For $0 \leqslant b, c \leqslant 100$, we have
$$
0 \leqslant 200-b-c \leqslant 200 \text {. }
$$

Therefore, $S$ contains $101^{2}$ elements.
Next, we prove: No element in $S$ is a multiple of another.

Assume $2^{200-b-c} 3^{b} 167^{c}$ is a multiple of $2^{200-i-j} 3^{i} 167^{j}$. Then
$$
\left\{\begin{array}{l}
200-b-c \geqslant 200-i-j, \\
b \geqslant i, \\
c \geqslant j,
\end{array}\right.
$$

which implies $b=i, c=j$.
Hence, no element in $S$ is a multiple of another.
We further prove by contradiction: The set $S$ can contain at most $101^{2}$ elements.
Suppose $U$ is a subset of $T$ with more than $101^{2}$ elements.
Since there are only $101^{2}$ distinct pairs $(b, c)$, by the pigeonhole principle, there must be two elements
$$
u_{1}=2^{a_{1}} 3^{b_{1}} 167^{c_{1}}, u_{2}=2^{a_{2}} 3^{b_{2}} 167^{c_{2}}
$$

such that $b_{1}=b_{2}, c_{1}=c_{2}, a_{1} \neq a_{2}$.
Thus, when $a_{1}>a_{2}$, $u_{1}$ is a multiple of $u_{2}$; when $a_{1}<a_{2}$, $u_{2}$ is a multiple of $u_{1}$.
Therefore, the subset $U$ does not satisfy the condition.
So, $S$ can contain at most $101^{2}$ elements.