We start with a 6-element subset $A$ of $\{1,\dots ,14\}$. What we'd like to prove is that there always exist two subsets of $A$, $A_1$ and $A_2$ such that: $$A_1 \cup A_2 = A$$ $$A_1 \neq A_2$$ $$\sum_{k\in A_1} k = \sum_{k \in A_2} k $$ That is, we want to split $A$ into two sets, possibly overlapping, whose sum of elements is equal.
I know this proof smells like pigeonhole principle, but how do I approach it? Furthermore how could I generalize it to k-element subsets of $\{1, \dots ,n\}$? When is it possible and when not?
You are right, the pigeon principle is key.
There are $2^6-2=62$ non trivial subsets of a subset of cardinality $6$. Each of these subsets will have a sum of elements between $1$ and $14+13+12+11+10=60$. So two of these subsets must have a sum that is equal.