20.3.7 ** Let $S$ be a non-empty subset of the set $\{1,2, \cdots, 108\}$, satisfying:
(i) For any two numbers $u, b$ in $S$ (which can be the same), there exists $c \in S$ such that $(a, c)=(b, c)=1$;
(ii) For any two numbers $a, b$ in $S$ (which can be the same), there exists $c' \in S$, different from $a, b$, such that $(a, c')>1, (b, c')>1$.
Find the maximum possible number of elements in $S$.
Parse the answer is 76.
Let $S$ be a set that meets the requirements, $|S| \geqslant 3$, and $p_{1}^{a} p_{2}^{\alpha} p_{3}^{\alpha} \in S$, where $p_{1}, p_{2}, p_{3}$ are 3 different primes, and $p_{1}1, (c_{3}, c_{2})>1$, so, the product of the smallest prime factors of $c_{1}$ and $c_{2}$ $\leqslant c_{3} \leqslant 108$, thus, $q \mid c_{1}$.

By $(c_{2}, c_{1})=1, (c_{2}, p_{1}^{\alpha} p_{2}^{\alpha} p_{3}^{\alpha})=1, \{p_{1}, p_{2}, p_{3}, q\}=\{2,3,5,7\}$, we know $c_{2}$ is a prime number greater than 10.

Again, by $(c_{3}, c_{2})>1$, we know $c_{2} \mid c_{3}$, and $11, (c_{5}, c_{4})>1$, so $c_{2} \mid c_{5}, c_{4} \mid c_{5}$, and $c_{2} \mid c_{3}$, $(c_{4}, c_{3})=1$, so $(c_{2}, c_{4})=1$, thus $c_{2} c_{4} \mid c_{5}$, but at this point, $c_{5} \geqslant c_{2} c_{4} \geqslant 11 \times 13 > 108$, a contradiction.

Therefore, $2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5, 2 \times 3 \times 7, 2^{2} \times 3 \times 7, 2 \times 5 \times 7, 3 \times 5 \times 7$ do not belong to $S$.
Next, we prove: the numbers $2 \times 3 \times 11, 2 \times 3 \times 13, 5 \times 7, 7$ do not all belong to $S$.
If not, by (i) there exist $d_{1}, d_{2} \in S$, such that $(2 \times 3 \times 11, d_{1})=1, (5 \times 7, d_{1})-1$, $(2 \times 3 \times 13, d_{2})=1, (5 \times 7, d_{2})=1$. So, $d_{1}$ and $d_{2}$ can only be prime numbers greater than 10. By (ii), $d_{1}=d_{2} \geqslant 17$. Again by (ii), there exists $d_{3} \in S$, such that $(7, d_{3})>1, (d_{2}, d_{3})>1$, so $7 d_{2} \mid d_{3}$, but $7 d_{2} \geqslant 7 \times 17=119$, a contradiction.

Furthermore, $1 \notin S$, and at most one prime number greater than 10 belongs to $S$. Thus, $|S| \leqslant 108-7-$ $1-23-1=76$.

On the other hand, take $S_{1}=\{1,2, \cdots, 108\} \backslash(\{1$ and primes greater than 11 $\} \cup\{2 \times 3 \times 11,2 \times$ $\left.3 \times 5,2^{2} \times 3 \times 5,2 \times 3^{2} \times 5,2 \times 3 \times 7,2^{2} \times 3 \times 7,2 \times 5 \times 7,3 \times 5 \times 7\right\}$, then $\left|S_{1}\right|=76$.
Next, we prove: $S_{1}$ satisfies conditions (i) and (ii).
In fact, for any $a, b \in S_{1}$, we discuss the following cases.
(1) If $a=b$, since at least one of 5, 7, 11 is not a divisor of $a$, (i) holds. For (ii), if $a$ is composite, take the smallest prime factor $p$ of $a$, then $p \in S_{1}$, and $(p, a)>1$; if $a$ is prime, then $a \leqslant$ 11. At this time $2 a \in S_{1}, (2 a, a)>1$.
(2) If $a \neq b$, if one of $a, b$ has 3 different prime factors, then this number can only be $2 \times 3 \times 13$ or $2 \times 3 \times 17$, without loss of generality, assume $a$ is such a number.
(1) $a=2 \times 3 \times 13, b \in S_{1}, b \neq a$, since one of 5, 7, 11 is not a divisor of $b$, let it

be $p$, then $(a, p)=(b, p)=1$, so (i) holds; for (ii), let $q_{1}$ be the smallest prime factor of $b$, then $2 q_{1} \leqslant 108, 3 q_{1} \leqslant 108$. If $b \neq 2 q_{1}$, then $2 q_{1} \in S_{1}$, and $(a, 2 q_{1})>1, (b, 2 q_{1})>1$, if $b=2 q_{1}$, then $3 q_{1} \in S_{1}$, and $(a, 3 q_{1})>1, (b, 3 q_{1})>1$. So (ii) holds.
(2) $a=2 \times 3 \times 17, b \in S_{1}, b \neq a$, the same discussion as (1) applies.
If $a, b$ both have at most two different prime factors, without loss of generality, assume $a1, (b, r_{1})>1$, (ii) holds;
If $r_{1}=r_{2}=a$, then take $u \in\{2,3\}$, such that $b \neq u a$, then $u a \in S_{1}, (u a, a)>1$, (ua, b) $>1$, (ii) holds;

If $r_{1} \neq r_{2}, r_{1} r_{2} \notin\{a, b\}$, then $r_{1} r_{2} \in S_{1}, (r_{1} r_{2}, a)>1, (r_{1} r_{2}, b)>1$, (ii) holds;

If $r_{1} \neq r_{2}, r_{1} r_{2}=a$, then take $u \in\{2,3\}$, such that $b \neq u r_{2}$, then $u r_{2} \in S_{1}$, $(u r_{2}, a)>1, (u r_{2}, b)>1$, (ii) holds;

If $r_{1} \neq r_{2}, r_{1} r_{2}=b$, then take $v \in\{2,3,5\}$, such that $a+v r_{1}, b \neq v r_{1}$, then $v r_{1} \in S_{1}$, $(v r_{1}, a)>1, (v r_{1}, b)>1$, (ii) holds.
In summary, $S$ has at most 76 elements.