I don't understand the quantifiers here. It sounds like it asks for the maximum of $k$ such that (for every directed graph $G$ on 2013 vertices in which every vertex has positive out-degree, for every vertex of $G$ we can reach at least $k$ vertices in at most two steps). But then the problem is basically trivial with answer $k = 2$. So, is something else meant?

JBL wrote: I don't understand the quantifiers here. It sounds like it asks for the maximum of $k$ such that (for every directed graph $G$ on 2013 vertices in which every vertex has positive out-degree, for every vertex of $G$ we can reach at least $k$ vertices in at most two steps). But then the problem is basically trivial with answer $k = 2$. So, is something else meant? I think it is true now.

I will present a solution to this rephrased statement, which I think correctly represents the problem. In a directed graph with $2013$ vertices, there is exactly one (directed) edge between any two vertices (this is called a tournament). For every vertex its out-degree is positive. From every vertex we can reach (at least) $k$ other vertices using at most two edges. Find the maximum value of $k$. I claim $\max k = 2012$, and I will call a tournament on $n$ vertices suitable if $\max k = n-1$. We can build a suitable tournament on $3$ vertices, say $a,b,c$, by taking as edges $\overrightarrow{ab}, \overrightarrow{bc}, \overrightarrow{ca}$. In general, if tournaments $T_a, T_b, T_c$ are suitable, then we can build a suitable tournament on $|T_a|+|T_b|+|T_c|$ vertices by taking the vertices of $T_a\cup T_b\cup T_c$, the edges from each of them, and all edges $\overrightarrow{ab}, \overrightarrow{bc}, \overrightarrow{ca}$, for all $a\in T_a, b\in T_b, c\in T_c$. Thus, we can build suitable tournaments with $3, 3+3+3=9, 9+3+3=15, 15+3+3=21$, etc., in general with $3+6m$ vertices. Since $2013 = 3 + 335\cdot 6$, we're done. EDIT. In fact, if $T$ is suitable, we can build a suitable tournament on $|T| + 2$ vertices by taking the vertices of $T$ plus two more, say $x$ and $y$, the edge $\overrightarrow{xy}$, the edges from $T$, and all edges $\overrightarrow{tx}, \overrightarrow{yt}$, for all $t\in T$. Thus, we can build suitable tournaments with $3, 3+1+1=5, 5+1+1=7$, etc., in general with odd number of vertices. Since $2013$ is odd, we're done.

Using mavropnevma's notation, it is fairly simple to prove that there are no suitable tournaments on $2$ or $4$ vertices, which may lead to the conjecture that there are none for an even number of vertices. So it is somewhat surprising that, for six vertices $v_1v_2v_3v_4v_5v_6$, the tournament consisting of the edges of the form $\overrightarrow{v_iv_{i+1}}$ for all $i$ (indices taken mod 6), $\overrightarrow{v_i v_{i+2}}$ for $i$ odd and $\overrightarrow{v_i v_{i-2}}$ for $i$ even, is suitable. In fact, the three edges of the form $a_{i} a_{i+3}$ are not necessary. Hence, there exists a suitable tournament on $n>1$ vertices for all $n \notin \{2, 4\}.$

This is the official English version of the problem: Between any two cities of a country consisting of $2013$ cities one-way flights are organized so that there is at least one departure from each city. Determine the maximal possible value of $k$ such that no matter how these flights are arranged there are $k$ cities reachable from any city of a country by using of at most two flights.