Pigeonhole principle, choosing 1-8 numbers out of 27 prove that for every 8 choosen numbers from 10 to 36 you can always make equalities.
number can be used once.
examples.
let say that the choosen numbers are 
10, 11, 12, 15, 18, 25, 32, 36
you can write
11+25=36 or 10+12+18=15+25.
i tried to prove for summation for every 2 numbers
$$\binom{8}{2}=28 $$
but there are 51 different summation of 2 differend numbers 
Lets say that the nombers are
a b c d e f g h
2 out of 8 is valid because the subsets (ab) and (ac) cannot be equal so they cannot be compared 
But 3 of 8 can be (abc) and (ade) and they can be equal so it is not valid
So i can use only use 2 out of 8 
Baybe i am in the wrong direction. 
that's the closest i got
 A: Let $S\subset\{10,11,12,\ldots,36\}$ be an arbitray subset of eight integers. The set $S$ itself has $2^8=256$ subsets. Of these $255$ are non-empty. For any subset $U\subset S$, $U\neq\emptyset$, let's denote by $\Sigma(U)$ the sum of elements of $U$. The smallest possible value of $\Sigma(U)$ is clearly
$$
\Sigma(\{10\})=10,
$$
and the largest possible value of $\Sigma(U)$ is
$$
\Sigma(\{36,35,34,\ldots,29\})=4\cdot65=260.
$$
These cannot occur for subsets of the same set $S$, but we don't need to worry about that. No matter what the set $S$ is, the $255$ numbers $\Sigma(U),U\subseteq S, U\neq\emptyset$, are all integers in the range $[10,260]$. That range has $251$ integers, so by the pigeonhole principle there exists two distinct subsets $U_1,U_2\subseteq S$ with the property that
$$
\Sigma(U_1)=\Sigma(U_2).
$$
Getting warmer, but let's not forget that a single element of $S$ can only be used once. In other words, the two sets $U_1,U_2$ should be disjoint. But this does not really form an obstacle for us. For if $V=U_1\cap U_2$ is non-empty, then, denoting $U_1'=U_1\setminus V$ and $U_2'=U_2\setminus V$, we see that
$$\Sigma(U_i)=\Sigma(U_i')+\Sigma(V)$$ for $i=1,2$.
Therefore
$$
\Sigma(U_1')=\Sigma(U_1)-\Sigma(V)=\Sigma(U_2)-\Sigma(V)=\Sigma(U_2').
$$
Thus the disjoint subsets $U_1'$ and $U_2'$ have equal set sums. Because obviously neither of $U_1,U_2$ can be a subset of the other, both $U_1'$ and $U_2'$ are non-empty.
