Question

Update: Let $S$ be the set. Let $X$ be the set of prime divisors of $S$. Without loss of generality we can assume that $X=2,3,...,p_k$ (all primes up to $p_k$). Indeed, we can always replace a bigger prime from $X$ by a smaller prime that does not without increasing the constant $C$. Now every element in $S$ $2^{l_1}...p_k^{l_k}$ corresponds to a vector $(l_1,...,l_k)$ in ${\mathbb Z}^k$. Let $\bar S$ be the set of all these vectors corresponding to numbers from $S$. Consider the partial component-wise order on ${\mathbb Z}^k$ (this makes the grid ${\mathbb Z}^k$ into a lattice (with intersection and join). Let $u_1,...,u_s$ be all the maximal vectors in $\bar S$ with respect to this partial order. We can assume that with every $x\in S$, $S$ contains all divisors of $X$. Therefore for every $u_i$, $\bar S$ contains all $v\le u$. These $v$'s form a parallelepiped $U_i$. The number of points in the union of all the parallelepipeds $U_i$ is $n$, the number of elements in $S$. Now we need to take any LCM of three $u_i$, and compare it with $k$. All the examples so far are such that there is only one maximal $u_i$ in $\bar S$. I think the only hope to prove that $C$ vanishes is to consider the case when there are many maximal vectors in $\bar S$. This is also the way to show that $C$ has a non-trivial lower bound.