erdos wrote: ... can be uniquely written as the sum of finitely many distinct elements of $A$.

I wanted to erase my post like two seconds ago . I forgot about periods, that's all .

One more try . I believe the answer to be negative. Assume $\{a_n\}_n=A$, and that $a_j>a_j$. We then have $a_i-a_j=\sum a_k$, and in order for $a_j+\sum a_k$ not to be a different representation of $a_i$ as a sum of finitely many distinct elements of $A$, $a_j$ must be one of the $a_k$'s which means that $a_i\ge 2a_j$. This happens whenever we have $a_i>a_j$, so we can order our elements $a_1\ge 2a_2\ge 4a_3\ge 8a_4\ge\ldots$ . Now, assume there is a $k$ s.t. $a_k>2a_{k+1}$. Take a rational $t$ strictly less that $a_k$ but still larger than $2a_{k+1}$. Then $t>a_{k+1}+a_{k+2}+\ldots$, so $t$ ahs no representation as the sum of finitely many distinct elements of $A$. This shows that for all $k\ge 1$, we must, in fact, have $a_k=2a_{k+1}$. This is impossible, because, for instance, $\frac{a_1}3$ cannot be represented as the sum of finitely many $a_k$'s. I hope it's correct this time.