If $a \in \mathbb{N}$, $b \in \mathbb{Z}_{\geq 0}$, and $A$ is the set of all numbers of the form $am + b$ (with $m$ a positive integer), then
\begin{equation*}
\sum_{n \in A} \frac{n}{2^n} = \sum_m \frac{am + b}{2^{am + b}} = \left(\frac{1}{2^b}\right)\left(\frac{a2^a}{(2^a - 1)^2}\right) + \left(\frac{b}{2^b}\right)\left(\frac{1}{2^a - 1}\right).
\end{equation*}
In particular, this number is rational. To see this, first observe that
\begin{equation*}
\sum_m \frac{am + b}{2^{am + b}} = \frac{1}{2^b}\sum_m \frac{am + b}{2^{am}}
\end{equation*}
\begin{equation*}
= \frac{1}{2^b}\sum_m \frac{am}{2^{am}} + \frac{1}{2^b}\sum_m \frac{b}{2^{am}}
\end{equation*}
\begin{equation*}
= \left(\frac{1}{2^b}\right)\sum_m \frac{am}{2^{am}} + \left(\frac{b}{2^b}\right)\left(\frac{1}{2^a - 1}\right).
\end{equation*}
Now all that remains is to prove that
\begin{equation*}
\sum_m \frac{am}{2^{am}} = \frac{a2^a}{(2^a - 1)^2}.
\end{equation*}
To see this, recall that
\begin{equation*}
\frac{1}{1 - x^a} = \sum_m x^{am}
\end{equation*}
for all $x \in (-1, 1)$. The desired result is obtained by differentiating with respect to $x$, multiplying by $x$, and then setting $x = \frac{1}{2}$.
(Note that this argument for $A = \mathbb{N}$ is already well-known; see, for example, this StackExchange post.)