Prove or disprove that in an 8-element subsets of $\{1,2…,30\}$ there must exist two $4$-element subsets that sum to the same number. How can I show that for any set of $8$ distinct positive integers not exceeding $30$, there must exist two distinct $4$-elements subsets that same up to the same number?
I tried using pigeon hole principle, but i still don't get it.
There are $$\binom {8}4=70$$ four-elements subsets of an $8$-element set. 
The least possible sum is $1+2+3+4=10$ and the greatest possible sum is $27+28+29+30=114$. Hence, there are $105$ sums. 
I have no idea how to continue because the number of possible integer sums is greater than the number of four-element subsets.
The $4$-element subsets are not necessarily non-overlapping.
Edit: 
For example, from $X=\{1,3,9,11,15,20,24,29\}$ , we can choose two different subsets $\{1,3,15,24\}$ and $\{3,9,11,20\}$ because they both sum up to $43$.

The statement is false.
Take for example a subset with 7 odd numbers and 1 even number.
Then we divide this subset into two 4-element subsets. One of them will have 4 odd numbers whose sum will be an even number while the other will have 1 even number and 3 odd numbers which will add up to an odd number.
Example: 1,3,5,7,9,11,13,14
Total sum is 63 and if we were to form 2 subsets, sum in each would not be a whole number.
