Example 4 (Canadian Mathematics Competition) Let the set $T$ be the set of all positive divisors of $2004^{100}$. Find the maximum possible number of elements in a subset $S$ of $T$, where no element of $S$ is a multiple of another.
Analyzing that if a number is a multiple of another number, then in the standard factorization, the exponent of each prime factor is not less than the corresponding prime factor's exponent of the other number, we seek a solution by standardly decomposing 2004.

Solution: Let $a, b, c$ be non-negative integers. Since $2004=2^{2} \cdot 3 \cdot 167$, we set
Clearly, $S$ is a subset of $T$. Since $b, c$ each have 101 possible values, $S$ contains $101^{2}=10201$ elements.

First, we prove that $S$ satisfies the condition. If some element $2^{100-b-c} 3^{b} 167^{c}$ is a multiple of another element $2^{100-i-j} 3^{i} 167^{j}$, then
$$
\left\{\begin{array}{l}
200 - b - c \geqslant 200 - i - j \\
b \geqslant i \\
c \geqslant j
\end{array} \Rightarrow \left\{\begin{array}{l}
b=i \\
c=j
\end{array}\right.\right. \text{, a contradiction. }
$$

Thus, $S$ satisfies the condition.
Second, we prove that the number of elements is at most 10201. If the set $U$ is a subset of set $T$ and has more than $101^{2}=10201$ elements, then the pairs $(b, c)$ have at most $101^{2}=10201$ pairs, so there must exist two elements $2^{i} 3^{b} 167^{c}$ and $2^{j} 3^{b} 167^{c}$, one of which is a multiple of the other, then $U$ does not satisfy the condition.