Numbers which are divisible by the sum of their prime factors.

We call a set of prime numbers good if the sum of its elements has no prime divisor outside of $A$.

Your problem is clearly equivalent to proving there are good sets of arbitrarily large size. We prove that for every $n$, a good set of size $n$ or $n+1$ exists:

Consider the set $A=\{p_1,p_2,\dots p_n\}$ of the first $n$ primes and let $q$ be their sum. Notice that $q\leq n p_n$, so if $A$ is not good then $q=pm$ with $m<n$ and $p$ a prime larger than $p_n$. The set $\{p_1,p_2,\dots, p_n,p\}$ is therefore good, since the sum of its elements is $p(m+1)$, and since $m+1\leq n$ clearly all of its prime factors are among $\{p_1,p_2,\dots, p_n,p\}$