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.
On
Here's a proof by contradiction. Call the tree $T$. For some $V \subseteq V(T)$, $T - V$ is the graph obtained by removing the vertices $V$ from $T$.
Suppose that your statement is false.
Take the "best" vertex $x$. That is, removing $x$ leaves a component $C$ of size strictly greater than $n/2$, but $C$ is as small as possible among possible choices of $x$.
Since $|V(C)| > n/2$, the graph $T - V(C)$ has at most $n/2$ vertices. Consider the unique neighbor $y$ of $x$ on $C$. Then removing $y$ divides $T$ into $T - V(C)$ and $C - \{y\}$. Because $T - V(C)$ has size at most $n/2$ and $C - \{y\}$ has size smaller than $C$, $y$ is a better choice than $x$, a contradiction.
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.