is there some information missing? See the attached $K_4$

No, there is no information missing. In your example the buttom-right vertex has the requested property, it has a red path to any other vertex.

how could I miss that

We'll prove it by induction. Show that it is true for $n=3$ vertices (trivial) and assume it is true for $n-1$. We'll say that a point that satisfies the given condition is a 'powerful point'. Take $G$ a graph with $n$ vertices. Suppose it has no powerful point. For any vertex $v$, the subgraph $G-v$ has $n-1$ vertices and thus has a powerful point. Then there's a powerful point for $G-v_1$, for $G-v_2$, ..., for $G-v_n$. If two of these were the same, this vertex would be a powerful point for $G$. So they must all be distinct. Thus each vertex has monochromatic paths to exactly $n-2$ other vertices; for each vertex $v$, let $f(v)$ be the vertex it doesn't reach. Consider the sequence $v, f(v), f(f(v)), \ldots$. Eventually some vertex must appear twice. Consider the first time this happens. If this vertex is not $v$ itself, then it must have two distinct 'ancestors' (points that don't reach it), which is absurd (since each vertex has exactly one other vertex that doesn't reach it). Then the first vertex to appear again must be $v$, and we must have a cycle. However, if the cycle contains fewer than $n$ vertices, then there must be a powerful point in this subgraph, which contradicts the fact that each vertex can't reach the next one in the cycle. Thus the cycle has $n$ vertices, that is, it is the whole graph. So order the vertices according to their position in the cycle as $v_1, v_2, \ldots, v_n$, where $f(v_i)=v_{i+1}$. Now we will draw more edges than there originally were. If there's a red path from $v$ to $u$, draw the red edge $vu$; the same goes for blue paths (of course, if the path from $v$ to $u$ is just the edge $vu$, we don't draw it again). Is it possible for two vertices $v, u$ to have both a blue edge $vu$ and a red edge $vu$? No, because then $v$ would reach all the points that $u$ reaches, and in particular $v$ would reach $f(v)$, absurd. It is easy to see, then, that each two vertices $v, u$ are joined by two edges $vu, uv$, each of which can be red or blue, unless $u=f(v)$ or vice-versa. Now, for each vertex $v$, consider the number of red edges $vx$. Clearly it is at most $n-2$ and at least $0$. So there must be two vertices with the same number of red edges $v, u$. Now, if $vu$ is red, then $v$ has at least as many red edges as $u$ does; in particular, for $u$ to have the same number of red edges, $uv$ must be red too. It follows that $vu, uv$ are either both red or both blue. Assume that they're both red (the other case is analogous). Now take $f(v), f(u)$. If $vf(u)$ was red, then there would be a red path $uvf(u)$ from $u$ to $f(u)$, absurd. Then $vf(u)$ is blue. Analogously $uf(v)$ is blue too. Now, if $f(v)v$ were blue, then there would be a blue path $uf(v)v$, from $u$ to $v$; then there would be a blue $uv$ edge too, absurd. So $f(v)v, f(u)u$ are red. Finally, if $f(v)f(u)$ were blue, then there would be a blue path $uf(v)f(u)$ from $u$ to $f(u)$, absurd. So $f(v)f(u), f(u)f(v)$ are red. By induction, $f(f(u)), f(f(v))$ are joined by red edges too, and so on. It follows, also, that $f(f(v))f(v)$ is red, and so on; by induction, $v_{i+1}v_i$ is red for all $i$. But then we can go from any vertex to any other vertex through a red path on the cycle!

What's a tournament?

A tournoment is a directed graph where between each two vertices there is exactly one directed edge. That is exactly one of these two edges are in the graph : $uv$ and $vu$

Nima Ahmadi Pour wrote: A tournoment is a directed graph where between each two vertices there is exactly one directed edge. That is exactly one of these two edges are in the graph : $uv$ and $vu$ Thank you.

Severius wrote: If this vertex is not $v$ itself, then it must have two distinct 'ancestors' (points that don't reach it) Why is this so ?

Severius wrote: So order the vertices according to their position in the cycle as $v_1, v_2, \ldots, v_n$, where $f(v_i)=v_{i+1}$. Now we will draw more edges than there originally were. If there's a red path from $v$ to $u$, draw the red edge $vu$; the same goes for blue paths (of course, if the path from $v$ to $u$ is just the edge $vu$, we don't draw it again). Is it possible for two vertices $v, u$ to have both a blue edge $vu$ and a red edge $vu$? No, because then $v$ would reach all the points that $u$ reaches, and in particular $v$ would reach $f(v)$, absurd. It is easy to see, then, that each two vertices $v, u$ are joined by two edges $vu, uv$, each of which can be red or blue, unless $u=f(v)$ or vice-versa. Now, for each vertex $v$, consider the number of red edges $vx$. Clearly it is at most $n-2$ and at least $0$. So there must be two vertices with the same number of red edges $v, u$. Now, if $vu$ is red, then $v$ has at least as many red edges as $u$ does; in particular, for $u$ to have the same number of red edges, $uv$ must be red too. It follows that $vu, uv$ are either both red or both blue. Assume that they're both red (the other case is analogous). Now take $f(v), f(u)$. If $vf(u)$ was red, then there would be a red path $uvf(u)$ from $u$ to $f(u)$, absurd. Then $vf(u)$ is blue. Analogously $uf(v)$ is blue too. Now, if $f(v)v$ were blue, then there would be a blue path $uf(v)v$, from $u$ to $v$; then there would be a blue $uv$ edge too, absurd. So $f(v)v, f(u)u$ are red. Finally, if $f(v)f(u)$ were blue, then there would be a blue path $uf(v)f(u)$ from $u$ to $f(u)$, absurd. So $f(v)f(u), f(u)f(v)$ are red. By induction, $f(f(u)), f(f(v))$ are joined by red edges too, and so on. It follows, also, that $f(f(v))f(v)$ is red, and so on; by induction, $v_{i+1}v_i$ is red for all $i$. But then we can go from any vertex to any other vertex through a red path on the cycle! Can't we simplify this second half a bit? If $v_n\to v_{n-1}\to\cdots\to v_1\to v_n$ is a monochromatic cycle, we're obviously done. Otherwise, there exists $i$ such that $v_{i+1}v_i$ and $v_iv_{i-1}$ are colored differently, say red and blue respectively. Now consider the monochromatic path from $v_{i-1}$ to $v_{i+1}\ne v_i=f(v_{i-1})$. If it's red, then we get a red path from $v_{i-1}$ to $v_i=f(v_{i-1})$, and if it's blue, then we get a blue path from $v_i$ to $v_{i+1}=f(v_i)$; either way yields a contradiction.

I had a slightly different proof than the ones I've seen so far. We will prove it by induction on the number of vertices $n$. Clearly for $n=2,3$ it is true, now suppose it is true for $n-1$. First label the graph $V=\{v_1,v_2,...v_n\}$. We say that a vertex $u$ "reaches" $v$ if there is a monochromatic directed path from $u$ to $v$ For each vertex $v_i$ we can apply the induction hypothesis on $V\setminus v_i$ and get that there is some vertex $v_j$ such that $v_j$ cannot reach $v_i$ but can reach every other vertex. Call this vertex the "king" for that set. Also for each set of $n-1$ a different king must be chosen otherwise if some vertex is chosen as the king twice it can reach every vertex so we are done. Hence each vertex in the graph can reach every vertex but one, and this choice is unique for different vertices. Although the start is the same I did not reach the same cycle idea as in the solution above. Instead I looked at $n=4$ and tried to generalize the contradiction I got there to all $n$, which worked well. Once you reach this point the idea is mostly about getting two paths of different colors between two points. Now there are two cases to consider: 1. There is some vertex with both a red and blue directed path coming from it Let's call that vertex $A$ and suppose that $A$ cannot reach $C$. However $A$ can reach every other vertex. Let $R=\{r_1,r_2,...,r_s\}$ be the vertices which $A$ can reach with a red path and $B=\{b_1,b_2,...,b_t\}$ the vertices $A$ can reach with a blue path. Now some vertex must not be able to reach $A$, suppose it is blue and call it $b_j$. Then $b_j$ can reach all vertices in $B$. Also by the induction hypothesis some vertex $r_i$ in $R$ can reach every vertex in $R$, so it can't reach some vertex $b_k$ in $B$. Now $r_iA$ must be red otherwise $r_iAb_k$ would be a blue path from $r_i$ to $b_k$. This means that $b_jr_i$ must be blue otherwise $b_jr_iA$ would be a red path from $b_j$ to $A$. But now if $r_iC$ is red, $Ar_iC$ is a red path from $A$ to $C$, and if it is blue, $Ab_jr_iC$ is a blue path from $A$ to $C$, contradiction. 2. Every vertex has only one color path from it to the vertices it can reach Suppose some vertex $A$ has red paths from it to $R=\{r_1,r_2,...,r_m\}$, and cannot reach some vertex $C$. Then $r_1C,r_2C,...,r_mC$ are all blue. Hence all paths coming from $r_1,r_2,...,r_m$ are blue. Suppose $r_i$ cannot reach $A$. Since $n\ge 4$, we have $m\ge 2$ so there is some other vertex $r_j$ which can reach $A$, but then $r_ir_jA$ is a blue path from $r_i$ to $A$, contradiction. So we are done.

Hopefully this works...

what a great problem for a directed graph $G$ let such $v$ vertex be the president of $G$ let $n$ be smallest integer such that there exists a graph $G$ doesn't satisfy the problem condition with $n$ vertices if we hide $v$ the let $f(v)$ be the president of $G - v$ the main claim: $v_1 \longrightarrow v_2\longrightarrow \dots v_n$ is a directed cycle proof: suppose that there are $u,v$ such that $w=f(v)=f(u)$ then there exist a colored path from $w$ to every vertex in $G-v \cup G-u =G$ contradiction so WLOG let $v_i=f^{i-1}(v_1)$ $\blacksquare$ now it's obvious that there isn't any colored path from $f(v)$ in $v$ so the edges $v_1v_2,v_2v_3 \dots v_nv_1$ aren't monochromatic WLOG let $v_1v_2$ be red and $v_2v_3$ be white consider the colored path $\vec{v_3v_1}$ if it is red then there's a red path $\vec{v_3v_2}$ contradiction if it's white then there's a white path $\vec{v_2v_1}$ contradiction and we win

Sands, B., Sauer, N. and Woodrow, R., 1982. On monochromatic paths in edge-coloured digraphs. Journal of Combinatorial Theory, Series B, 33(3), pp.271-275. Corollary 2.

We induct on $|G|$, where the base cases $|G| = 1, 2, 3$ are trivial. For the inductive step, assume for contradiction that $G$ does not satisfy the desired property. Note that for every vertex $v \in G$, there exists some $f(v) \in G$ which has a directed edge pointing to each other vertex $G \setminus v$ per the inductive hypothesis. If $f(v) \rightarrow v$ then we are done, so assume henceforth that $v \rightarrow f(v)$ for all $v \in G$. Repeating $f$ iteratively, we have \[ v \to f(v) \to \dots \to f^{k-1}(v) \to f^k(v) \]is a cycle $\mathcal{C}$ for some $k$. Assume $\mathcal{C}$ that it is not monochromatic, or else we are done. We now direct our attention to the vertices $v$, $f(v)$, and $f^2(v)$, as looking at them is sufficient to finish. WLOG $v \to f(v)$ is colored red and $f(v) \to f^2(v)$ is colored blue; consider the following casework. If $v \sim f^2(v)$ is colored red, then $f^2(v)$ works as the desired vertex, and if $v \sim f^2(v)$ is colored blue, then $f(v)$ works. The inductive step is complete, finishing.

The result for one colored tournaments seems like folklore. Anyways, nice problem. We show this inductively. The base case of $n = 1, 2$ is immediate. FTSOC suppose not for $n \ge 3$. Let $V_i$ be the such vertex of the tournament when removing the vertex $i$, which follows inductively. If $V_i = V_j$, then contradiction follows. Thus the map $i \mapsto V_i$ is injective and thus bijective. Else, all $V_i$ are distinct. Claim: There exists no monochromatic directed path from $V_i$ to $i$. Proof. If there did, then this leads to contradiction because then $V_i$ would also lead to $i$. $\blacksquare$ As such, it follows that the edge from $i$ to $V_i$ is directed towards $V_i$. Claim: The path from $V_i$ to $V_{V_i}$ and the path from $i$ to $V_i$ are the same color. Proof. We have a monochromatic path $V_{V_{i}} \to i$, as well as monochromatic direct paths from $i \to V_i$ and $V_i \to V_{V_{i}}$. Then neither of the paths $V_{V_i} \to i \to V_i$ and $V_i \to V_{V_i} \to i$ can be monochromatic. It thus follows that $i \to V_i$ and $V_i \to V_{V_{i}}$ both have the opposite color of $V_{V_{i}} \to i$ and thus have the same color. $\blacksquare$ However, now the chain $i \to V_i \to V_{V_{i}} \to \dots$ is eventually cyclic and monochromatic, contradicting the existence of a monochromatic path from $V_i$ to $i$.

The problem is true for $n=1,2$, we can use strong induction. Hopefully this is right because the idea seems different from the other posts. Let $w$ be a vertex. Then $G\setminus w$ has a vertex $v$ such that $v$ can reach all other vertices with a monochromatic path. If $v$ points to $w$ then $v$ can reach every vertex in $G$ with a monochromatic path, so suppose $w$ points to $v$. WLOG this edge is red. Let $B$ be the set of vertices that $v$ can reach with blue edges, and let $R$ be the same for red edges. Then by the inductive hypothesis, there is a vertex $x$ in $B\cup w$ such that $x$ can reach all other vertices in $B\cup w$ with monochromatic paths. $x=w$ or $x$ can reach $w$ with a red path. Then $x$ can reach all vertices in $B\cup w$ with a monochromatic path. There is a red path from $x$ to $w$ to $v$, so $x$ can use a red path to reach all vertices in $R$ as well. $x\neq w$ but $x$ can reach $w$ with a blue path. Since $v$ can reach $x$ with a blue path, then $v$ can reach $w$ with a blue path. $v$ can reach all other vertices with a monochromatic path, as desired.