Existence of infinite set of positive integers s.t sum of reciprocals is rational and set of primes dividing an element is infinite

The set $$ A=\{1\times 2, 2\times 3, ...,n\times (n+1),...\}$$

is such a set.

Note that $$\sum _1^\infty \frac{1}{n(n+1)}=1$$ and every prime number $p$ divides $p(p+1)$ which is an element of the set $A$.

We can take $(a_i)$ to be an increasing sequence of primes: Let $a_1 = 2$, and choose the following prime to be the smallest one for which the sum of reciprocals is strictly less than $1$.

The sum of reciprocals will converge to $1$; this is because $\sum_p \frac1p = \infty$:

Fact. When $x_i > 0$, $x_i \to 0$ and $\sum x_i = \infty$, then for every positive real number $y$ there exists a subsequence $x_{i_k}$ for which $\sum_k x_{i_k} = y$.

Proof. Construct the subsequence as above. Concretely, let $S_0 = 0$ and define recursively for $k \geq 0$, $i_{k+1} = \min\{i : S_k + x_i < y \}$ and $S_{k+1} = S_k + x_{i_{k+1}}$. Suppose $\sum_k x_{i_k} = y - \epsilon$. Then all $x_i$ which are less than $\epsilon$ appear in the subsequence. Thus $$\sum_{x_i < \epsilon} x_i \leq y - \epsilon < \infty,$$ a contradiction.

Let $f=(f_1,f_2): \mathbb{N} \rightarrow \mathbb{N}^2$ be a bijection.

Define $a_n=(2^{3+f_1(n)}+1)^{f_2(n)}$.

By Bang’s theorem (https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem), if $n > m > 3$, then there exists some prime $p$ dividing $2^{2n}-1$ but neither $2^{2m}-1$ nor $2^n-1$, thus $p|2^n+1$ but not $p|2^m+1$.

As a consequence, no integer can be both a power of $2^m+1$ and $2^n+1$, hence the $a_n$ are pairwise distinct.

Besides, $$\sum_n{a_n^{-1}}=\sum_{n \geq 1}{\sum_{m \geq 1}{(1+2^{n+3})^{-m}}}=\sum_{n \geq 1}{2^{-3-n}} \in \mathbb{Q}.$$