I want to prove the statement in the title: for any tree on $n$ vertices, it is possible to delete a vertex such that the deletion leaves connected components with at most $n/2$ vertices each.
I drew some example trees (e.g. a star) and noticed that vertices of high degree should do the trick. For trees that vertices of low degree, then one of the "central" vertices work. However, I can't find a way to determine when to pick between "central" vertices and vertices of high degree.
Best Answer
Each edge $\{u,v\}$ of the tree partitions $V$ into two nonempty parts. Draw an arrow on the edge $\{u,v\}$ pointing in the direction of $v$ if the $v$-part of $V$ contains more than half of the vertices. Then no edge gets two arrows. Starting at a leaf proceed along arrows as long as possible. The resulting path has to end at some vertex $v$, because otherwise it would contain a loop. The vertex $v$ has no outgoing arrow. This implies that the removal of $v$ does not create any connected component with $>|V|/2$ vertices.