The maximum number of elements is $72$.

A positive integer not greater than $100$ can be written as
$$n = 2^{a_1} \cdot 3^{a_2} \cdot 5^{a_3} \cdot 7^{a_4} \cdot 11^{a_5} \cdot q,$$
where $q$ is a positive integer and not divisible by $2$, $3$, $5$, $7$ and $11$, and $a_1, a_2, a_3, a_4$ and $a_5$ are nonnegative integers.

We pick out those positive integers $n$ with just one or two nonzero among $a_1, a_2, a_3, a_4$ and $a_5$ to form set $S$. In this case, $S$ contains $50$ even numbers ($2, 4, \ldots, 98$ and $100$) except the following seven: $2 \times 3 \times 5$, $2^2 \times 3 \times 5$, $2 \times 3^2 \times 5$, $2 \times 3 \times 7$, $2^2 \times 3 \times 7$, $2 \times 5 \times 7$ and $2 \times 3 \times 11$; $17$ odd numbers that are multiples of three (i.e. $3 \times 1, 3 \times 3, \ldots, 3 \times 33$), $7$ odd numbers with the least prime divisor $5$ (i.e. $5 \times 1, 5 \times 5, 5 \times 7, 5 \times 11, 5 \times 13, 5 \times 17$ and $5 \times 19$), $4$ odd numbers with the least prime divisor $7$ (i.e. $7 \times 1, 7 \times 7, 7 \times 11$ and $7 \times 13$), and the prime number $11$.

Consequently, $S$ contains $(50-7)+17+7+4+1=72$ numbers totally.

In what follows, we will prove that $S$ constructed above satisfies the given condition.

Obviously, it satisfies condition (1).

For condition (2), we note that, at most, four prime divisors among $2, 3, 5, 7$ and $11$ will occur in $[a, b]$. We write the prime which does not occur as $p$. Obviously, $p \in S$ and
$$(p, a) \le (p, [a, b]) = 1, \\ (p, b) \le (p, [a, b]) = 1.$$
Hence, we take $c = p$.

For condition (3), we take the least prime divisor of $a$ as $p$ and the one of $b$ as $q$ when $(a, b) = 1$. It is easy to see that $p \ne q$ and $p, q \in \{2, 3, 5, 7, 11\}$. Hence $pq \in S$, and
$$(pq, a) \ge p > 1 \text{ and } (pq, b) \ge q > 1.$$
Being coprime to each other for $a$ and $b$ ensures that $pq$ is different from $a$ and $b$. Thus we take $d = pq$.

When $(a, b) = e > 1$, we take $p$ as the least prime divisor of $e$, and $q$ as the smallest prime number satisfying $q \nmid [a, b]$. It is easy to see that $p \ne q$, and $p$ and $q \in \{2, 3, 5, 7, 11\}$. Hence $pq \in S$, and
$$(pq, a) \ge (p, a) = p > 1, \\ (pq, b) \ge (p, b) = p > 1.$$
$q \nmid [a, b]$ ensures that $pq$ is different from $a$ and $b$. Thus, we take $d = pq$.

In what follows, we prove that the number of elements in $S$, which satisfies the conditions described in the problem, will not be greater than $72$.

Obviously, $1 \notin S$. For arbitrary two prime numbers $p$ and $q$ which are both greater than $10$, since the least number which is not prime to neither $p$ nor $q$ is $pq$, it must be greater than $100$. So we know, according to condition (3), that there is at most one among $21$ prime numbers between $10$ and $100$ ($11, 13, \ldots, 89, 97$), occurring in $S$. We write the set consisting of all natural numbers not greater than $100$ except $1$ and the above-mentioned $21$ prime numbers as $T$, and there are $78$ numbers in the set. We can conclude that there are at least $7$ numbers in $T$ are not in $S$. Thus $S$ contains at most $78 - 7 + 1 = 72$ elements.

i. When a prime number $p$ is greater than $10$ and belongs to $S$,

every number in $S$ can only have $2, 3, 5, 7$ and $p$ as its least prime divisor. By condition (2), we have the following conclusions.

① If $7p \in S$, because $\{2 \times 3 \times 5, 2^2 \times 3 \times 5, 2 \times 3^2 \times 5, 7p\}$ contains all the least prime divisors, we know from condition (2) that $2 \times 3 \times 5, 2^2 \times 3 \times 5$ and $2 \times 3^2 \times 5$ do not belong to $S$. If $7p \notin S$, noting $2 \times 7p > 100$, but $p \in S$, so from condition (2) we know that $7 \times 1, 7 \times 7, 7 \times 11$ and $7 \times 13$ do not belong to $S$.

② If $5p \in S$, then $2 \times 3 \times 7$ and $2^2 \times 3 \times 7$ do not belong to $S$. If $5p \notin S$, then $5 \times 1$ and $5 \times 5$ do not belong to $S$.

③ $2 \times 5 \times 7$ and $3p$ do not belong to $S$ at the same time.

④ $2 \times 3p$ and $5 \times 7$ do not belong to $S$ at the same time.

⑤ If $5p, 7p \notin S$, then $5 \times 7 \notin S$.

When $p = 11$ or $13$, from ①, ②, ③ and ④, we can get at least $3, 2, 1$ and $1$ numbers in $T$ respectively which do not belong to $S$, and in total $7$ numbers. When $p = 17$ or $19$, from ①, ② and ③, we can get at least $4, 2$ and $1$ numbers in $T$ respectively, which do not belong to $S$ and in total $7$ numbers. When $p > 20$, from ①, ② and ③, there are at least $4, 2$ and $1$ numbers in $T$ respectively, which do not belong to $S$ and in total $7$ numbers also.

ii. If there is no prime number greater than $10$ belonging to $S$, then the least prime numbers in $S$ can only be $2, 3, 5$ and $7$. Hence, each of the following $7$ pairs of numbers can not belong to $S$ at the same time:
$$(3, 2 \times 5 \times 7), (5, 2 \times 3 \times 7), (7, 2 \times 3 \times 5), (2 \times 3, 5 \times 7), (2 \times 5, 3 \times 7), (2 \times 7, 3 \times 5), (2^2 \times 7, 3^2 \times 5).$$
Thus, there are at least $7$ numbers in $T$ that are not in $S$.

Consequently, the answer for this problem is $72$.