Answer: 61

Small Hint:

If $S$ had six elements, compare the variance of its $64$ subset sums with that of $64$ consecutive integers

Big Hint:

After bounding the size of $S,$ inspect the five-element subsets whose sums exceed the candidate

Solution:

First, $S$ has at most five elements. If it had six elements $s_1,\ldots,s_6,$ then its $64$ subset sums would all be distinct: equality between two subset sums, after cancelling their common elements, would violate the given condition.

Choose a subset uniformly at random and let $X$ be its sum. Then $$\begin{gathered} \operatorname{Var}(X) =\frac14\sum_{i=1}^6s_i^2\\ {}\leq\frac14(10^2+11^2+\cdots+15^2)\\ {}=\frac{955}{4}. \end{gathered}$$ On the other hand, $X$ is uniform on $64$ distinct integers. The least possible variance for $64$ distinct integers occurs when they are consecutive, and is $$\frac{64^2-1}{12}=\frac{1365}{4},$$ a contradiction.

A set with at most four elements has sum at most $12+13+14+15=54.$ There are only seven five-element subsets of $\{1,\ldots,15\}$ whose sums are at least $62.$ Each fails, as witnessed by the following equal sums:

$\{8,12,13,14,15\}:$ $13+14=12+15.$ $\{9,11,13,14,15\}:$ $11+13=9+15.$ $\{9,12,13,14,15\}:$ $13+14=12+15.$

$\{10,11,12,14,15\}:$ $11+14=10+15.$ $\{10,11,13,14,15\}:$ $11+13=10+14.$

$\{10,12,13,14,15\}:$ $12+13=10+15.$ $\{11,12,13,14,15\}:$ $12+13=11+14.$

Thus the answer is at most $61.$

The set $\{8,11,13,14,15\}$ attains $61.$ Its $32$ subset sums, in order, are $$\begin{gathered} 0,8,11,13,14,15,19,21,\\ 22,23,24,25,26,27,28,29,\\ 32,33,34,35,36,37,38,39,\\ 40,42,46,47,48,50,53,61, \end{gathered}$$ all distinct. Equal sums from arbitrary subsets would, after deleting their intersection, give equal sums from disjoint subsets, so this verifies the required property.