Example 6 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 $a, b$ in $S$, there exists an element $c$ in $S$ such that the greatest common divisor of $a$ and $c$ is 1, and the greatest common divisor of $b$ and $c$ is also 1;
(3) For any two different elements $a, b$ in $S$, there exists an element $d$ in $S$ different from $a, b$ such that the greatest common divisor of $a$ and $d$ is greater than 1, and the greatest common divisor of $b$ and $d$ is also greater than 1. (2003 China Mathematical Olympiad Problem)
Analysis and Solution: Represent the positive integer $n \leq 100$ as $n=2^{\alpha_{1}} 3^{\alpha_{2}} 5^{\alpha_{3}} 7^{\alpha_{4}} 11^{\alpha_{5}} \cdot m$, where $m$ is a positive integer not divisible by $2, 3, 5, 7, 11$, and $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are natural numbers. Select those positive integers for which exactly 1 or 2 of $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-zero to form the set $S$. That is, $S$ includes: 50 even numbers $2, 4, \cdots, 100$ but excluding $2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5, 2 \times 3 \times 7, 2^{2} \times 3 \times 7, 2 \times 5 \times 7, 2 \times 3 \times 11$ (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. Clearly, $|S| = (50 - 7) + 17 + 7 + 4 + 1 = 72$. We prove that this constructed $S$ meets the conditions.

Condition (1) is obviously satisfied.
For condition (2), for any two different elements $a, b \in S$, the prime factors of $[a, b]$ include at most 4 of the numbers $2, 3, 5, 7, 11$. Let the missing one be $p$, then $p \in S$, and $(p, a) \leq (p, [a, b]) = 1$, $(p, b) \leq (p, [a, b]) = 1$. Thus, taking $c = p$ works.

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 $p, q \in \{2, 3, 5, 7, 11\}$, so $pq \in S$, and $(pq, a) \geq p > 1$, $(pq, b) \geq q > 1$. The fact that $a$ and $b$ are coprime ensures that $pq$ is different from $a$ and $b$, so taking $c = pq$ works. When $(a, b) = e > 1$, take the smallest prime factor $p$ of $e$ and the smallest prime $q$ that does not divide $[a, b]$. Clearly, $p \neq q$, and $p, q \in \{2, 3, 5, 7, 11\}$, so $pq \in S$, and $(pq, a) \geq (p, a) = p > 1$, $(pq, b) \geq (p, b) = p > 1$. The fact that $q$ does not divide $[a, b]$ ensures that $pq$ is different from $a$ and $b$, so taking $d = pq$ works.

We now prove that for any set $S$ satisfying the conditions, $|S| \leq 72$.
Clearly, $1 \notin S$; for any two primes $p, q > 10$, the smallest number not coprime with both $p$ and $q$ is $pq$, and $pq > 100$, so according to condition (3), at most one of the 21 primes in $[10, 100]$: 11, 13, ..., 89, 97 can be in $S$. Let the set of the remaining 78 positive integers not exceeding 100, excluding 1 and these 21 primes, be $T$. We prove that at least 7 numbers in $T$ are not in $S$, thus $|S| \leq 78 - 7 + 1 = 72$.

Actually, if some prime $p > 10$ is in $S$, the smallest prime factor of each number in $S$ can only be 2, 3, 5, 7, or $p$. Combining condition (2), we know: (i) If $7p \in S$, since $2 \times 3 \times 5, 2^2 \times 3 \times 5, 2 \times 3^2 \times 5$ and $7p$ include 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 $7p \notin S$, and $2 \times 7p > 100$, and $p \in S$, then by condition (3), $7 \times 1, 7 \times 7, 7 \times 11, 7 \times 13 \notin S$. (ii) If $5p \in S$, then $2 \times 3 \times 7, 2^2 \times 3 \times 7 \notin S$. If $5p \notin S$, then $5 \times 1, 5 \times 5 \notin S$. (iii) $2 \times 5 \times 7$ and $3p$ cannot both be in $S$. (iv) $2 \times 3p$ and $5 \times 7$ cannot both be in $S$. (v) If $5p, 7p \notin S$, then $5 \times 7 \notin S$. Thus, when $p = 11$ or 13, by (i), (ii), (iii), (iv), at least 3, 2, 1, 1 numbers in $T$ are not in $S$, totaling 7. When $p = 17$ or 19, by (i), (ii), (iii), at least 4, 2, 1 numbers in $T$ are not in $S$, totaling 7. When $p > 20$, by (i), (ii), (iii), at least 4, 2, 1 numbers in $T$ are not in $S$, totaling 7. Therefore, the conclusion holds.

When no prime $p > 10$ is in $S$, the smallest prime factor of each number in $S$ can only be 2, 3, 5, 7. In this case, at least one number in each of the 7 pairs of sets $\{3, 2 \times 5 \times 7\}, \{5, 2 \times 3 \times 7\}, \{7, 2 \times 3 \times 5\}, \{2 \times 3, 5 \times 7\}, \{2 \times 5, 3 \times 7\}, \{2 \times 7, 3 \times 5\}, \{2^2 \times 7, 3^2 \times 5\}$ is not in $S$, so the conclusion holds.

In summary, $|S|_{\max} = 72$.