A recently-deleted question asked to show that if $G(n)$ is the graph on $[n]=\{1,\dots,n\}$ where $x$ and $y$ are adjacent if $x|y$ or $y|x$, then $G(n)$ is planar. This is false if $n\geq150$, for the subgraph induced by $\{2,3,5,90,120,150\}$ is isomorphic to $K_{3,3}$.
My question is, what is the largest value of $n$ such that $G(n)$ is planar? Of course, various software packages would make short work of this, but I'd be interested in a verbal argument, if someone can give one.
It can be no greater than $15$, as $\{1,2,4,8,16\}$ gives a $K_5$. I was able to draw the graph for $14$, shown partially completed below. $1-6$ has to go inside so that $3-12$ can go outside, but I think it is clear how to complete it. If you try to do it on $15$ you fail because merging $5,10,15$ makes a $K_{3,3}$ with $\{1,2,3\}$ and $\{6,12,5-10-15\}$. Even if you swap $7,14$ with $5,10$ so $15$ can go in the region to connect to both $3$ and $5$ it blocks the $1-6$ connection.