There are $ \binom{10}{2}=45$ pairs of elements from the drawn set (let's call it $ S$). Since there are 36 possible non-negative differences of numbers between 1 and 37, there are 9 coincidences of the form $ a-b=c-d$ with $ a,b,c,d\in S$. If all four numbers in such a coincidence are different, then (adding $ b$,$ d$ to both sides) we are done with the proof already. So let us assume that all of these coincidences are of the form $ a-b=c-a$ for some $ a,b,c\in S$ Let $ M$ be the maximum and $ m$ be the minimum numbers from $ S$. Notice that $ c-M$ and $ m-b$ cannot yield positive differences when $ c,b\in S$, hence numbers $ a$ in $ a-b=c-d$ must be all from the remaining 8 elements of $ S$. Since there are 9 coincidences, for some two of them we have the same $ a$, i.e. $ a-b_1=c_1-a$ and $ a-b_2=c_2-a$. Substracting both equalities we have $ b_2+c_2=c_1+b_1$. It is east to notice that $ b_1,b_2,c_1,c_2$ must all be different. best regards, Wojtek