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Suppose $A$ is a set of $8$ distinct positive integers, and $x \le 30$ for all $x \in A$. Show that there must be two distinct subsets of $A$ with $4$ elements whose elements when added up give the same sum.

I have tried to solve this problem as follows:

Observe that for $A_1 = \{1,2,3,4,27,28,29,30\}$, the least possible sum is $1+2+3+4=10$, and the largest sum is $27+28+29+30=114$. Thus, there are $105$ possible sums. However, the number of subsets of $A$ with 4 elements are $C(8,4) = 70$.

And I got stuck there. The idea is to shows that the number of possible sums is less than the number of subsets and prove it with pigeon hole principle. Then for all of other possible set $A$, the number of possible sums won't be more than the number of sums in $A_1$. However, I am still not sure about the general way of calculating all of the possible sums. What's the correct way of doing that?

Octav K.
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    This was posted 20 hours ago (no solutions yet) - https://math.stackexchange.com/questions/3465258/prove-or-disprove-that-in-an-8-element-subsets-of-1-2-30-there-must-exist – Calvin Lin Dec 07 '19 at 01:56
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    @CalvinLin This probably means that there is a source of this problem, which both posters don't give - or it's just pure coincidence? – WhatsUp Dec 07 '19 at 02:09
  • In case this is from an ongoing contest, please note the contest problem policy. – joriki Dec 07 '19 at 07:14
  • I am not really sure if it is from a contest, but its an assignment from a university discrete math course. As there is already a post on the question, should I take mine down? – Octav K. Dec 07 '19 at 08:44
  • @OctavK.: That would make sense, yes. It currently can't be closed as a duplicate by others because the other question doesn't have an upvoted or accepted answer. – joriki Dec 07 '19 at 09:11

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