6) Let $S$ be a non-empty subset of the set $1.2, \cdots, 108$, satisfying:
(i) For any two numbers $a, b$ in $S$ (which can be the same), there exists $c \in S$ such that $(a \cdot c)=(b \cdot c)=1$;
(ii) For any two numbers $a, b$ in $S$ (which can be the same), there exists $c' \in S, c'$ not equal to $f'(a, b)$, such that $(a \cdot c')>1, (b \cdot c')>1$.
Find the maximum possible number of elements in $S$.
The answer is 76.
Let $S$ be a set that meets the requirements. $|S| \geq 3$. And $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha} \in S$, here $p_{1}, p_{2}, p_{3}$ are 3 different primes, and $p_{1}1 .\left(c_{3} \cdot c_{2}\right)>1$. So $, c_{1}, c_{2}$'s smallest prime factor product $\leqslant c_{3} \leqslant$ 108, thus, $q \mid c$.

By $\left(c_{2} \cdot c_{1}\right)=1,\left(c_{2} \cdot p_{1}^{\alpha_{1}} p_{2}^{\alpha} \cdot p_{3}^{\alpha_{i}}\right)=1,\left\{p_{1} \cdot p_{2} \cdot p_{3} \cdot q\right\}=$ 12.3.5.7 we know $\mathrm{C}$ is a prime greater than $\mathrm{f}: 10$.

Again, by $\left(c_{3}, c_{2}\right)>1$, we know $c_{2} \mid c_{3}$, and $11,\left(c_{5} \cdot c_{4}\right)>1$. Therefore, $c_{2} \mid c_{5}$, $c_{4} \mid c_{3}$, and $c_{2} \mid c_{3},\left(c_{4}, c_{3}\right)=1$. So $\left(c_{2}, c_{4}\right)=1$. Thus $c_{2} c_{4} \mid c_{5}$. But at this point, $c_{3} \geqslant c_{2} c_{+} \geqslant 11 \times 1.3>108$. 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 $\left(2 \times 3 \times 11, d_{1}\right)=1$, $\left(5 \times 7, d_{1}\right)=1,\left(2 \times 3 \times 13, d_{2}\right)=1,\left(5 \times 7, d_{2}\right)=1$. So, $d_{1}, d_{2}$ can only be primes greater than 10. By (ii) we know $d_{1}=d_{2} \geqslant 17$. Again, by (ii) there exists $d_{3} \in S$ such that $\left(7, d_{3}\right)>1,\left(d_{2}, d_{3}\right)>1$, so $7 d_{2} \mid d_{3}$, but $7 d_{2} \geqslant 7 \times 17=119$. Contradiction.

Further, $1 \notin S, H$. Among the primes greater than 10, at most one belongs to $S$. Thus, $S \mid \leqslant 108-7-1-23-1=76$.

On the other hand, take $S_{1}=\{1,2, \cdots, 108\} \backslash(\{1$ and primes greater than 11 $\} \cup$ $2 \times 3 \times 11, 2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5, 2 \times 3 \times 7, 2^{2} \times 3 \times 7, 2 \times 5 \times 7, 3 \times 5 \times 7\}$ ), then $\left|S_{1}\right|=76$.
Next, we prove: $S_{1}$ satisfies conditions (i) and (ii).
Fact 1. For any $a, b \in S_{1}$, we discuss as follows.
Case one $a=b$. By $\mathrm{f}: 5, 7, 11$ at least one of these numbers is not a divisor of $a$. Hence (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 $2 a \in S_{1}$, $(2 a, a)>1$.
Case two $a \neq b$.
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 $(a, p)=(b, p)=1$, hence (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 $\left(a, 2 q_{1}\right)>1, \left(b, 2 q_{1}\right)>1$; if $b=2 q_{1}$, then $3 q_{1} \in S_{1}$, and $\left(a, 3 q_{1}\right)>1, \left(b, 3 q_{1}\right)>1$. 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 $a1, \left(b, r_{1}\right)>1$, (ii) holds;
If $r_{1}=r_{2}=a$, then take $u \in \{2,3\}$, such that $b \neq u a$, then $u a \in S_{1}$,
$(u a \cdot a)>1, (u a, b)>1$, (ii) holds;
If $r_{1} \neq r_{2}, r_{1} r_{2} \notin \{a, b\}$, then $r_{1} r_{2} \in S_{1}, \left(r_{1} r_{2}, a\right)>1$, $\left(r_{1} r_{2}, b\right)>1$, (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 $u r_{2} \in S_{1}, \left(u r_{2}, a\right)>1, \left(u r_{2}, b\right)>1$, (ii) holds; if $r_{1} \neq r_{2}, r_{1} r_{2}=b$, then take $v \in \{2,3,5\}$, such that $a \neq v r_{1}, b \neq v r_{1}$, then $v r_{1} \in S_{1}, \left(v r_{1}, a\right)>1, \left(v r_{1}, b\right)>1$, (ii) holds.
In summary, we know that $S$ has at most 76 elements.