Parse the maximum value of $|S|$ as 72.
Represent each positive integer $n$ not exceeding 100 as
$$
n=2^{\alpha_{1}} \cdot 3^{\alpha_{2}} \cdot 5^{\alpha_{3}} \cdot 7^{\alpha_{4}} \cdot 11^{\alpha_{5}} \cdot q,
$$

where $q$ is a positive integer not divisible by $2, 3, 5, 7, 11$, and $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-negative integers.
We select the positive integers that satisfy the condition "exactly 1 or 2 of $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-zero" to form the set $S$. Thus, $S$ includes 50 even numbers $2, 4, \cdots, 98, 100$, but excludes $2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5, 2 \times 3 \times 7, 2^{2} \times 3 \times 7, 2 \times 5 \times 7, 2 \times 3 \times 11$—these 7 numbers; 17 odd multiples of 3: $3 \times 1, 3 \times 3, \cdots, 3 \times 33$; 7 numbers with the smallest prime factor 5: $5 \times 1, 5 \times 5, 5 \times 7, 5 \times 11, 5 \times 13, 5 \times 17, 5 \times 19$; 4 numbers with the smallest prime factor 7: $7 \times 1, 7 \times 7, 7 \times 11, 7 \times 13$; and the prime number 11. Therefore, $S$ contains a total of $(50-7) + 17 + 7 + 4 + 1 = 72$ numbers.
Below, we prove that the constructed $S$ satisfies the given conditions.
Condition (1) is obviously satisfied.
For condition (2), note that among the prime factors of $[a, b]$, at most 4 of $2, 3, 5, 7, 11$ appear. Let the missing number be $p$. Clearly, $p \in S$, and
$$
\begin{array}{l}
(p, a) \leqslant (p, [a, b]) = 1, \\
(p, b) \leqslant (p, [a, b]) = 1.
\end{array}
$$

Thus, taking $c = p$ suffices.
For condition (3), when $(a, b) = 1$, take the smallest prime factor $p$ of $a$ and the smallest prime factor $q$ of $b$. Clearly, $p \neq q$, and $p, q \in \{2, 3, 5, 7, 11\}$. Thus, $pq \in S$, and
$$
(pq, a) \geqslant p > 1, \quad (pq, b) \geqslant q > 1.
$$
The fact that $a$ and $b$ are coprime ensures that $pq$ is different from $a$ and $b$. Thus, taking $d = pq$ suffices.
When $(a, b) = e > 1$, take $p$ as the smallest prime factor of $e$ and $q$ as the smallest prime not dividing $[a, b]$. Clearly, $p \neq q$, and $p, q \in \{2, 3, 5, 7, 11\}$. Thus, $pq \in S$, and
$$
\begin{array}{l}
(pq, a) \geqslant (p, a) = p > 1, \\
(pq, b) \geqslant (p, b) = p > 1.
\end{array}
$$
The fact that $q \nmid [a, b]$ ensures that $pq$ is different from $a$ and $b$. Thus, taking $d = pq$ suffices.
Next, we prove that the number of elements in any set $S$ satisfying the given conditions does not exceed 72.
Clearly, $1 \notin S$. For any two primes $p, q$ greater than 10, since the smallest number not coprime with both $p$ and $q$ is $pq$, which is greater than 100, by condition (3), at most one of the 21 primes between 10 and 100, i.e., $11, 13, \cdots, 89, 97$, can be in $S$. Let the set of the remaining 78 natural numbers not exceeding 100, excluding 1 and these 21 primes, be denoted as $T$. We claim that at least 7 numbers in $T$ are not in $S$, thus $S$ can have at most $78 - 7 + 1 = 72$ elements.
(i) When there is a prime $p > 10$ in $S$, the smallest prime factor of all numbers in $S$ can only be $2, 3, 5, 7$, and $p$. Using condition (2), we can derive the following conclusions:
(1) If $7p \in S$, since $2 \times 3 \times 5, 2^2 \times 3 \times 5, 2 \times 3^2 \times 5$ cover all the smallest prime factors, by condition (2), $2 \times 3 \times 5, 2^2 \times 3 \times 5, 2 \times 3^2 \times 5 \notin S$; if $7p \notin S$, note that $2 \times 7p > 100$, and $p \in S$, so by condition (3), $7 \times 1, 7 \times 7, 7 \times 11, 7 \times 13 \notin S$.
(2) If $5p \in S$, then $2 \times 3 \times 7, 2^2 \times 3 \times 7 \notin S$; if $5p \notin S$, then $5 \times 1, 5 \times 5 \notin S$.
(3) $2 \times 5 \times 7$ and $3p$ cannot both be in $S$.
(4) $2 \times 3p$ and $5 \times 7$ cannot both be in $S$.
(5) If $5p, 7p \notin S$, then $5 \times 7 \notin S$:
When $p = 11$ or 13, by (1), (2), (3), and (4), we can respectively conclude that at least $3, 2, 1, 1$ numbers in $T$ are not in $S$, totaling 7 numbers; when $p = 17$ or 19, by (1), (2), and (3), we can respectively conclude that at least $4, 2, 1$ numbers in $T$ are not in $S$, totaling 7 numbers; when $p > 20$, by (1), (2), and (3), we can respectively conclude that at least $4, 2, 1$ numbers in $T$ are not in $S$, totaling 7 numbers.
(ii) If no prime greater than 10 is in $S$, then the smallest prime factor of numbers in $S$ can only be $2, 3, 5, 7$. Thus, the following 7 pairs of numbers cannot both be in $S$:
$$
\begin{array}{c}
(3, 2 \times 5 \times 7), (5, 2 \times 3 \times 7), (7, 2 \times 3 \times 5), (2 \times 3, 5 \times 7), \\
(2 \times 5, 3 \times 7), (2 \times 7, 3 \times 5), (2^2 \times 7, 3^2 \times 5).
\end{array}
$$

Thus, at least 7 numbers in $T$ are not in $S$.
In summary, the maximum value of $|S|$ is 72.
20.3 .9 * Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two different elements $a$ and $b$ in $S$, there exists an element $c$ in $S$ such that the greatest common divisor (gcd) of $a$ and $c$ is 1, and the gcd of $b$ and $c$ is also 1;
(3) For any two different elements $a$ and $b$ in $S$, there exists an element $d$ in $S$ different from $a$ and $b$ such that the gcd of $a$ and $d$ is greater than 1, and the gcd of $b$ and $d$ is also greater than 1.