We prove a stronger statement that already among subsets of $A$ with at most four elements there are two with equal sums of elements. Set $A$ has
$$\binom{7}{1} + \binom{7}{2} + \binom{7}{3} + \binom{7}{4} = 98$$
subsets with at most four elements. The sum of elements of each of these subsets is at least $1$ and at most $26 + 25 + 24 + 23 = 98$. Assume that among these subsets there are no two with equal sums of elements. Then every number between $1$ and $98$ is the sum of elements of exactly one subset of $A$ with at most four elements. In particular, $98$ is the sum of the subset $\{23, 24, 25, 26\} \subset A$. On the other hand, two-element subsets $\{23, 26\}$ and $\{24, 25\}$ have equal sums, which is a contradiction.