Our universe is the set of positive integers $M = \{1,2,3,4,5,6,\ldots, 30m\}$. There are $15m$ odd numbers; among them there are $3m$ divisible by $5$, so $S$ has $12m$ elements. Among them there are $4m$ divisible by $3$, hence $8m$ not divisible by $3$. For each number $a$ of these, consider the disjoint sets (chains of the divisibility poset) $C_a = \{3^{\alpha}a \mid \alpha=0,1,\ldots\} \cap S$, and $M_a = \max C_a$. The set $A = \{M_a \mid a \in S, 3 \nmid a\} \subset S$ is an antichain (so no element of it divides another), and $|A| = 8m$. On the other hand, any subset of $S$ with more than $8m$ elements will have two in a same $C_a$ (by pigeonhole principle), thus one divisible by the other. Therefore $k=8m+1$. (The proof is a verbatim application of Erdös' method).