Mathematica competitions in Croatia

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 $$\left(\genfrac{}{}{0ex}{}{7}{1}\right)+\left(\genfrac{}{}{0ex}{}{7}{2}\right)+\left(\genfrac{}{}{0ex}{}{7}{3}\right)+\left(\genfrac{}{}{0ex}{}{7}{4}\right)=98$$ 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.