Sorry the title isn't complete, but my title was over 150 characters, so here is the question:
What is the least possible value of $n$ in a set of the form ${1, 2, ..., n}$ such that it can be partitioned into $2$ subsets in which one must have two elements $a$ and $b$ so that $ab$ is divisible by $a+b$.
This question is probably easiest done by guess and check and just grinding through until you get it, but I wanted to know if there was a more quick and safe way to do it. Thanks!
$x+y~|~ xy\Rightarrow \exists k~~s.t. (x+y)k=xy~~or, ~~(x-k)(y-k)=k^2$. Now we want $x\neq y$. So wlog assume $x<y$, then $x-k<k<y-k$ such that $(x-k)(y-k)=k^2$. $k$ cannot be $1$, because this leads to $x=y=2$. If $k=2$, then $k^2=4=1\times4=(3-2)\times(6-2)$. So smallest such $n$ for which $\{1,\cdots, n\}$ contains two numbers, such that their product is divisible by their sum, is $6$.