Solution:
Assume, for a contradiction, that there exists such a set.

We first prove that if $a \in A$, then $A \cap \left(\frac{a}{2}, a\right) = \varnothing$. To do this suppose the contrary, i.e. there exists $a'$ in $A$ and $a > a' > \frac{a}{2}$. Then the number $a - a' < \frac{a}{2}$ can be represented as a sum of one or finitely many different numbers from $A$. Since each of these numbers is less than $\frac{a}{2}$, the number $a = a' + (a - a')$ has two different representations of the required type (as $a$ and as $a'$ plus the numbers of the representation of $a - a'$), a contradiction.

In particular, it follows from the above that in every interval $\left[\frac{1}{2^i}, \frac{1}{2^{i-1}}\right)$, $i = 1, 2, \ldots$, there is at most one element of $A$. Since the set $A$ is infinite (otherwise we can obtain only a finite number of sums of different numbers from $A$) it easily follows that the numbers of $A$ can be ordered in an infinite sequence $a_1, a_2, \ldots$, which satisfies $a_i \geq 2 a_{i+1}$ for every $i$. If this inequality is strict for some $i$, then
$$s = \sum_{i=2}^{\infty} a_i < \sum_{i=2}^{\infty} \frac{a_1}{2^{i-1}} = a_1$$
This shows that the numbers from the interval $(s, a_1)$ cannot be represented as a sum of one or finitely many different numbers from $A$.

Therefore $a_{i+1} = \frac{a_1}{2^i}$ for every $i$. Now it is easy to see that only the numbers of the form $a_1 \frac{m}{2^n}$ can be represented as a sum of one or a finite number of different numbers from $A$. Thus any rational number with an odd denominator which is coprime with the denominator of $a_1$ cannot be represented as required. This is a contradiction which completes the proof.