If a $10$-element subset whose sum is 155 is removed from $\{1,2,...,30\}$, then it's always possible to split the remaining set into equal parts. In other words, I'd like to prove that, given the set $\{1,2,...,30\}$, if we designate a $10$-element subset whose sum is 155 (for example $\{1,2,3,4,5,26,27,28,29,30\}$), then the remaining 20 elements can always be split into two 10-element subsets whose sums are equal (in our example, $\{6,8,10,12,14,17,19,21,23,25\}$ and $\{7,9,11,13,15,16,18,20,22,24\}$).
So far all I have is that the sum $1+2+...+30 = 465 = 3\cdot 155$, so all three subsets in the problem will have a sum of $155$. After that, I can't think of anything. Could someone please give me a hint to push me in the right direction? It doesn't seem like a very complicated problem, so that would be more than enough.

So first pair up the numbers as $r, 31-r$ and note that any such pair can replace any other without changing the sum of a subset. 
You are given a set of ten elements adding to $155$. Consider constructing a second set as made up of elements $31-r$ where $r$ is in the first set. The sum of elements in both is then $310=2\times 155$, but there may be elements in the second set which are also in the first. Show that you can use the properties of pairs (and that there are enough pairs) to eliminate duplicates from the second set without changing its sum. The remaining elements form the third set, will have the right sum (and will, by this method, consist of five pairs).
Have deleted full answer because you wanted a hint.
