I know it is possible to show the following by using the pigeon hole principle. Given a set $A_{100} = \{ 1, 2, 3, ..., 100\}$, if a subset $S$ of $A$ is randomly chosen with $|S| = 10$, there exist two disjoint subsets of $S$ having the same sum.
But when I think about what is the maximum $k \in \mathbb{N}$ that whenever a random subset $S$ of $A_{k} = \{ 1, 2, 3, ..., k\}$, with $|S| = 10$ is chosen, there exist two disjoint subsets of $S$ having the same sum.
I think this is the same as finding the least $k'$ that all subsets of $S \subset A_k$, with $|S| = 10$, have a different sum. ($k' = k+1$)
I notice that the possible sum of elements in $S$ is at least $\min (S)$ and at most $\sum_{s_i \in S} s_i$. Thererfore the range of possible sum is at most $\sum_{s_i \in S} s_i - \min(S)$.
In the case of $A_{k'}$, $\sum_{s_i \in S} s_i - \min(S) \leq (k'-8) + (k'-7) + \cdots + k' = 9k' - 36$. Then for $2^{10} = 1024$ of possible subset of $S$, and by pigeon hole principle, $9k' - 36 \geq 1024$ for no same sum being found. So $k'$ is at least 118.
But I don't know an efficient way to show whether such $S \subset A_{118}$ exists and so don't know how to process.
I think such $k'$ should exist as for $k'=512$, we can choose $S = \{1,2,4,8,16,32,64,128,256,512\}$ and all sums are different integers ranging from $0$ to $1023$.