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 $43$.

Let the elements of $X$ be $a_1<a_2<...<a_8$ and denote the seven successive differences by $d_i=a_{i+1}-a_i.$

Consider the subsets of size $4$ which contain either $2$ or $3$ elements of $\{a_5,a_6,a_7,a_8\}$. There are $$\begin{pmatrix}4\\1\\\end{pmatrix}\begin{pmatrix}4\\3\\\end{pmatrix}+\begin{pmatrix}4\\2\\\end{pmatrix}\begin{pmatrix}4\\2\\\end{pmatrix}=52$$ of these subsets and the possible sums of their elements range from $a_1+a_2+a_5+a_6$ to $a_4+a_6+a_7+a_8$. So, by the pigeon-hole principle, we are finished unless $$a_4+a_6+a_7+a_8-(a_1+a_2+a_5+a_6)+1\ge 52$$ $$\text {i.e.} 2(a_8-a_1)\ge51+d_1+d_4+d_7.$$ Since $a_8-a_1\le 29$ we must have $d_1+d_4+d_7\le7$. Using the observations given below, $d_1,d_4,d_7$ are all different and no two can add to the third and so $\{d_1,d_4,d_7\}=\{1,2,4\}$ and $\{a_1,a_{8}\}=\{1,30\}.$

Some observations about the $d_i$.

(1) Any two non-adjacent differences are unequal.

(2) Given three non-adjacent differences, none is the sum of the other two.

(3) Given two adjacent differences, the sum of these differences can replace one of the differences in observations (1) and (2). (We still require the 'combined difference' to be non-adjacent to the other differences involved.)

The proofs of these are all elementary and of the same form. As an example, suppose we have $d_2+d_3=d_5+d_7$, which is a combination of (2) and (3). Then $$a_4-a_2=a_6-a_5+a_8-a_7.$$ The sets $\{a_4,a_5,a_7\}$ and $\{a_2,a_6,a_8\}$ then have the same sum and $a_1$, say, can be added to each.

To return to the main proof where we know that the differences $\{d_1,d_4,d_7\}=\{1,2,4\}$.

Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $1$. Then, by the observations, $\{d,d+1\}\cap\{2,4,6\}$ is empty. So $d\ge7$.

Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $2$. Then, by the observations, $\{d,d+2\}\cap\{1,3,4,5\}$ is empty. So $d\ge6$.

Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $4$. Then, again by the observations, $\{d\}\cap\{1,2,3\}$ is empty. So $d\ge4$.

The sum of the differences (which is $29$) is now at least $(1+2+4)+(7+6+4)+d$, where $d$ is the 'other' difference adjacent to $d_4$. Therefore $d_4=4$ and the two differences adjacent to it (which cannot be equal) are $4$ and $5$. The differences adjacent to the differences of $1$ and $2$ are thus forced to be $7$ and $6$, respectively. Then $a_1+a_8=a_3+a_5$ and we are finished.

This is NOT a proof.

I have written a code and run it, and indeed, in every $8-$plet of different numbers among $\{1,2,\ldots,30\}$, there exist (at least) two different quadruplets with the same sum.

The most interesting however is that, this holds even when $n=30$ is replaced by $n=31, ,32,\ldots,40$. In the case for $n=41$ (and apparently for every number larger than $41$), such $8-$plets do exist. In particular, for $n=41$, there exist exactly $4$ such $8-$plets: $$ 1,\,2,\,3,\,11,\, 20,\, 35,\, 38,\, 41 \\ 1,\,2,\,3,\,20,\, 29,\, 35,\, 38,\, 41 \\ 1,\,4,\,7,\,13,\, 22,\, 39,\, 40,\, 41 \\ 1,\,4,\,7,\,22,\, 31,\, 39,\, 40,\, 41 $$