This is a question that I just wondered. I don't know if there's a good answer or not.
Given a complete graph of $n$ vertices. Each edge $ab$ is given a direction (either $a\rightarrow b$ or $b\rightarrow a$).
What is a maximum $k$ such that there always exist $k$ distinct vertices $a_1,a_2,\ldots,a_k$ such that we have the directed path $a_1\rightarrow a_2\rightarrow\ldots\rightarrow a_k$?
The answer is n. This is because every tournament (directed graph where there is exactly one directed edge between every pair of vertices) has Hamiltonian path, and this is what I'm about to show by induction:
Basis is trivial, so I'll head directly to induction step.
Let's suppose that every graph containing n-1 vertices has Hamiltonian path. Now, let's take graph G and choose some vertice labeled $v_0$. Now, by induction assumption we know that $G/ \{ v_0 \}$ contains hamiltonian path: $v_1, v_2, \dots v_{n-1}$. Now let i be such maximal index that there is edge from every $v_1, v_2,...v_{i}$ to $v_0$ (there also could be none, but this isn't a problem). This also means that there is edge from $v_0$ to $v_{i+1}$, otherwise i wouldn't be maximal. Finally, Hamiltonian path in graph $G$ is:
$v_1,v_2,\dots,v_{i},v_0,v_{i+1},\dots, v_n$
Q.E.D.