Example 7 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 numbers $a, b$ in $S$, there exists a number $c$ 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 numbers $a, b$ in $S$, there exists a number $d$ in $S$ such that the gcd of $a$ and $d$ is greater than 1, and the gcd of $b$ and $d$ is also greater than 1.
(2003 CMO Problem)
Let $S$ be a set that satisfies conditions (1) and (3). By condition (3), we know that $1 \notin S$. Let $p$ and $q$ be any two distinct primes greater than 10. If $p$ and $q$ both belong to $S$, the smallest number that is not coprime with both $p$ and $q$ is $pq$, which is greater than 100, contradicting condition (3). Therefore, among the 21 primes between 10 and 100, $11, 13, 17, \cdots, 89, 97$, at most one belongs to $S$. Let the set of the remaining 78 positive integers not exceeding 100, excluding 1 and these 21 primes, be denoted as $T$. We will discuss how many numbers in $T$ do not belong to $S$.

(i) When there is a prime $p$ greater than 10 that belongs to $S$, the smallest prime factor of all numbers in $S$ can only be $2, 3, 5, 7$, and $p$.
(1) If $7p \in S$, then $S$ does not contain $2 \times 3 \times 5, 2^2 \times 3 \times 5, 2 \times 3^2 \times 5$, otherwise, taking $a = 7p$ and $b$ as any of $2 \times 3 \times 5, 2^2 \times 3 \times 5, 2 \times 3^2 \times 5$, for any $c \in S$ different from $a$ and $b$, $c$ must contain at least one of the prime factors $2, 3, 5, 7, p$, thus at least one of $a$ and $b$ is not coprime with $c$, contradicting condition (2). If $7p \notin S$ and $p \in S$, then $1 \times 7, 7 \times 7, 7 \times 11, 7 \times 13 \notin S$, otherwise, taking $a = p$ and $b$ as any of $1 \times 7, 7 \times 7, 7 \times 11, 7 \times 13$, by condition (3) there exists $d \in S$ such that $a$ and $b$ are not coprime with $d$, hence $d$ is divisible by both 7 and $p$. Since $7p \notin S$, $d \geq 2 \times 7p > 100$, so $d \notin S$, a contradiction.
(2) If $5p \in S$, then by condition (2), $2 \times 3 \times 7, 2^2 \times 3 \times 7 \notin S$. If $5p \notin S$, noting that $2 \times 5p > 100$ and $p \in S$, by condition (3), $5 \times 1, 5 \times 5 \notin S$.
(3) By condition (2), $2 \times 5 \times 7$ and $3p$ cannot both belong to $S$.
(4) By condition (2), $2 \times 3p$ and $5 \times 7$ cannot both belong to $S$.
(5) If $5p, 7p \notin S$, noting that $2 \times 5p > 100$ and $2 \times 7p > 100$ and $p \in S$, by condition (3), $5 \times 7 \notin S$.

When $p = 11$ or 13, from (1), (2), (3), and (4), we can respectively conclude that at least 3, 2, 1, and 1 numbers in $T$ do not belong to $S$, totaling $3 + 2 + 1 + 1 = 7$ numbers not in $S$; when $p = 17$ or 19, $7p \notin S$. From (1), (2), and (3), we can respectively conclude that at least 4, 2, and 1 numbers in $T$ do not belong to $S$, totaling $4 + 2 + 1 = 7$ numbers not in $S$; when $p > 20$, from (1), (2), and (5), we can respectively conclude that at least 4, 2, and 1 numbers in $T$ do not belong to $S$, totaling $4 + 2 + 1 = 7$ numbers not in $S$.
(ii) If no prime greater than 10 belongs to $S$, then the smallest prime factor of the numbers in $S$ can only be $2, 3, 5, 7$. Thus, the following 7 pairs of numbers cannot both appear in $S$:
$$
(3, 2 \times 5 \times 7), (5, 2 \times 3 \times 7), (7, 2 \times 3 \times 5), (2 \times 3, 5 \times 7), (2 \times 5, 3 \times 7), (2 \times 7, 3 \times 5), (2^2 \times 7, 3^2 \times 5).
$$

Thus, at least 7 numbers in $T$ are not in $S$, so $S$ can contain at most $78 - 7 + 1 = 72$ numbers.
The following example shows that $S$ can contain 72 numbers.
First, $S$ includes 50 even numbers $2, 4, 6, \cdots, 98, 100$, except for $2 \times 3 \times 5, 2^2 \times 3 \times 5, 2 \times 3^2 \times 5, 2 \times 3 \times 7, 2^2 \times 3 \times 7, 2 \times 5 \times 7, 2 \times 3 \times 11$, a total of 7 numbers; 17 odd multiples of 3, $3 \times 1, 3 \times 3, 3 \times 5, \cdots, 3 \times 33$; 7 odd numbers with the smallest prime factor 5, $5 \times 1, 5 \times 7, \cdots, 5 \times 19$; 4 odd numbers with the smallest prime factor 7, $7 \times 1, 7 \times 7, 7 \times 11, 7 \times 13$; and the prime 11. Thus, $S$ contains a total of $(50 - 7) + 17 + 7 + 4 + 1 = 72$ numbers.
The following proves that the constructed $S$ satisfies the given conditions.
Condition (1) is clearly satisfied. For condition (2), note that for any $a, b \in S (a \neq b)$, the least common multiple of $a$ and $b$ has at most 4 prime factors from $2, 3, 5, 7, 11$ (otherwise, one of $a$ or $b$ would be greater than 100, a contradiction). Let the missing prime be $p$, clearly $p \in S$, and
$$
(p, a) \leq (p, [a, b]) = 1, (p, b) \leq (p, [a, b]) = 1.
$$

Thus, taking $c = p$ satisfies condition (2).
For condition (3), when $(a, b) = 1$, take the smallest prime factor of $a$ as $p$ and the smallest prime factor of $b$ as $q$. Clearly, $p \neq q$, and $p, q \in \{2, 3, 5, 7, 11\}$, so $pq \in S$, and
$$
(a, pq) \geq p > 1, (b, pq) \geq q > 1.
$$
The fact that $a$ and $b$ are coprime ensures that $pq$ is different from $a$ and $b$, thus taking $d = pq$ satisfies condition (3).
When $(a, b) = e > 1$, take the smallest prime factor of $e$ as $p$ and the smallest prime factor of $[a, b]$ as $q$. Clearly, $p = q$, and $p, q \in \{2, 3, 5, 7, 11\}$, so $pq \in S$, and
$$
(pq, a) \geq (p, q) = p > 1, (pq, b) \geq (p, b) = p > 1.
$$
The fact that $q \times [a, b]$ ensures that $pq$ is different from $a$ and $b$, thus taking $d = pq$ satisfies condition (3).
Therefore, the constructed set $S$ satisfies all three conditions.
In conclusion, the maximum number of elements in $S$ that satisfy the conditions is 72.