China $1996$ Number theory Problem 
Find the smallest positive integer $K$ such that every $K$ -element subset of $\{1,2, \ldots, 50\}$ contains two distinct elements $a, b$ such that $a+b$ divides $a b$

Let $c=g c d(a, b),$ so $a=c a_{1}$ and $b=c b_{1}$. 
Therefore, $c a_{1} b_{1}$ is divisible by $a_{1}+b_{1}$ 
Furthermore, since $g c d\left(a_{1}, b_{1}\right)=1,$ we see that $a_{1}+b_{1}$ is relatively prime to $a_{1}$ and $b_{1},$ s $\left(a_{1}+b_{1}\right) | c$
since $a+b \leq 99 \Rightarrow a_{1}+b_{1} \leq 9 .$ 
How $a+b \leq 99 \Rightarrow a_{1}+b_{1} \leq 9 .$ ???

Since $a_1+b_1|c$  then $(a_1+b_1)^2|c(a_1+b_1)=a+b$.
