For any set $A$ of numbers, let $sum(A)$ be the sum of the elements of $A$.
Consider the set $B = \{1,3,6,8,10,11\}$. There are 20 subsets of $B$ containing three elements, and their sums are:
$$\begin{align}
& sum(\{1,3,6\}) = 10 \\
& sum(\{1,3,8\}) = 12 \\
& sum(\{1,3,10\}) = 14 \\
& sum(\{1,3,11\}) = 15 \\
& sum(\{1,6,8\}) = 15 \\
& sum(\{1,6,10\}) = 17 \\
& sum(\{1,6,11\}) = 18 \\
& sum(\{1,8,10\}) = 19 \\
& sum(\{1,8,11\}) = 20 \\
& sum(\{1,10,11\}) = 22 \\
& sum(\{3,6,8\}) = 17 \\
& sum(\{3,6,10\}) = 19 \\
& sum(\{3,6,11\}) = 20 \\
& sum(\{3,8,10\}) = 21 \\
& sum(\{3,8,11\}) = 22 \\
& sum(\{3,10,11\}) = 24 \\
& sum(\{6,8,10\}) = 24 \\
& sum(\{6,8,11\}) = 25 \\
& sum(\{6,10,11\}) = 27 \\
& sum(\{8,10,11\}) = 29
\end{align}$$
Some of these sums occur more than once, others are unique. For a set $A$, let $U(A,k)$ be the set of unique sums of $k$-element subsets of $A$, in our example we find $U(B,3) = \{10,12,14,18,21,25,27,29\}$ and $sum(U(B,3)) = 156$.
Now consider the $100$-element set $S = \{1^2, 2^2, \ldots , {100}^2\}$. $S$ has $100\,891\,344\,545\,564\,193\,334\,812\,497\,256\;$ $50$-element subsets.
Determine the sum of all integers which are the sum of exactly one of the $50$-element subsets of $S$, i.e. find $sum(U(S,50))$.