In the mathoverflow question here the asymptotic growth of antichains in the divisibility poset ${\cal P}_n$ of the set of natural numbers $\{1,\ldots,n\}$ is considered. I have a somewhat dual question.

Consider a ``packing'' of disjoint chains $C_x$ where $$C_x=\{x,y_1 x, y_1 y_2 x, \ldots \} \in \{1,2,\ldots,n\}$$ with $y_i$ natural numbers greater than 1. Thus, $x$ is the minimal element in the chain $C_x$ and $C_{x'}\cap C_x=\emptyset$ whenever $x \neq x'.$ The cost of a chain $C_x$ is simply

$f({\cal C}_x)=1/x $

where $x$ is its minimal element. I want to find a collection of disjoint chains ${\cal M}=\{C_{x_1},\ldots,C_{x_m}\}$ which maximizes the cost $$f({\cal M})=\sum_{i=1}^m f(C_{x_i})=\sum_{i=1}^m \frac{1}{x_i}$$ for large $n$.

Edit: There is a side condition to the maximization, which I forgot to write, sorry. To be able to include both $x\neq x'$ in $\cal M$ they must not divide each other in addition to their LCM being strictly greater than $n$.

An upper bound of $\lceil \frac{n+1}{2}\rceil$ is known on the size of a collection obeying this LCM condition, see this previous MO question here. However, the size maximizing sets that are easy to find are essentially mostly integers from $[n/2,n]$ and the value of $f$ they give is constant (since it is $H_n-H_{n/2}$). However, it is known that for so called primitive sets in $\{1,\ldots,n\}$ considered by Behrend, Erdos and others (sets of integers satisfying the non-divisibility conditions) the value of $f$ is proved to be asymptotically $\frac{\log n}{\sqrt{\log \log n}}$.

It seems like some pruning of primitive sets by removing their members in $[2,\sqrt{n+1})\cap \mathbb{N}$ and pruning multiples of small primes we should get a collection with cost more than a constant.