Subset sum problem with many divisibility conditions
This problem can be solved in polynomial time using linear programming, and this is actually true for any partial order $(S,\le)$. By the way, we can prove by induction that for any finite partial order set  $(S,\le)$, there exists a finite set $S'\subseteq\mathbb{N}$ and a bijection $f:S\rightarrow S'$, such that for all $s_1,s_2\in S, s_1\le s_2 \Leftrightarrow f(s_1) | f(s_2)$.

Let $\mathcal{C}$ be the set formed by the chains in $S$. Remind that $C$ is a chain iff for all $v,v'$ in $C$, $v\le v'$ or $v'\le v$

Now create a boolean variable $x_v$ for each $v\in S$, and a boolean variable $y_C$ for each chain $C$. We can write the following linear program $(P)$ for our problem :
$$
\begin{split}
\text{Max} \displaystyle\sum\limits_{v\in S} x_v \\
\text{subject to} \displaystyle\sum\limits_{v\in C} &amp;x_v \le 1, \forall C\in\mathcal{C}\\
                                         &amp;x_v \in \{0,1\}, v\in S
\end{split}
$$

and its dual $(D)$  :

$$
\begin{split}
\text{Min} \displaystyle\sum\limits_{C\in \mathcal{C}} y_C\\
\text{subject to} \displaystyle\sum\limits_{C:v\in C} &amp;y_C \ge 1, \forall v\in S\\
                                         &amp;y_C \in \{0,1\}, C\in \mathcal{C}
\end{split}
$$

Then the problem of finding the minimum cover of an ordered set by chains is the dual of our problem. Dilworth's theorem states that 

There exists an antichain A, and a partition of the order into a family P of chains, such that the number of chains in the partition equals the cardinality of A

which means that the optimal solution of these two problems match : $Opt(P)=Opt(D)$ 

Let $(P^*)$ (resp. $(D^*)$) be the relaxation of $(P)$ (resp. $(D)$) i.e. the same linear program where all constraints $x_v\in\{0,1\}$ (resp. $y_C\in\{0,1\}$) are replaced by $x_v\in [0,1]$ (resp. $y_C\in [0,1]$). Let $Opt(P^*)$ and $Opt(D^*)$ be their optimal solutions. Since $\{0,1\}\subseteq [0,1]$ we have :
$$
Opt(P)\le Opt(P^*) \text{ and } Opt(D^*)\le Opt(D)
$$
and weak duality theorem establishes that $Opt(P^*)\le Opt(D^*)$ then by putting everything together we have :
$$
Opt(P)= Opt(P^*)=Opt(D^*)=Opt(D)
$$

Then, using Ellipsoid method, we can compute $Opt(P^*)$ ( $=Opt(P)$) in polynomial time. There are an exponential number of constraints but there exists a polynomial time separation oracle. Indeed given a solution $X$, we can enumerate all couples $s_1,s_2\in X$ and check if $s_1\le s_2$ or $s_2\le s_1$, and therefore decide in polynomial time whether $X$ is feasible or otherwise the constraint associated to the chain $\{v_1,v_2\}$ is violated.