Is it possible that everyone is a pseudo-winner in a tournament with 25 people? Is it possible, that everyone is a pseudo-winner in a tournament with 25 people?
(pseudo-winner means that either he won against everyone, or if he lost against someone, then he beated someone else, who beated the one who he lost to).
In the "language" of graph theory: Is it possible in the directed K(25) (all edges drawn with 25 vertices), that from all vertices, every vertices can be reached with a maximum of 2-long route.
I thought about this, and my answer is no. I tried to do the proving, starting with K(3), which is a triangle. Triangle works, but I don't know how to continue this until 25 vertices.
Maybe any easier solutions?
Thanks!
 A: We can prove that this is possible for any odd integer $2n+1$ by induction. 
For $n=1$ it is possible. 
If it is possible for $n\ge 1$ Let $G$ have $2(n+1)+1$ vertices. Let $u,v$ be some $2$ vertices and the others are $v_1,v_2,...,v_m$ where $m=2n+1$ Connect $v_1,...,v_m$ by induction. so that from ever vertex there is path of length at most $2$ to any other. Now draw directed edges from $u->v_i$ and $v_i->v$ for $i=1,...,m$ and $v->u$. This graph has all pseudo-winners and has $2(n+1)+1$ vertices thus induction is complete. 
Thus for $25$ it is possible aswell.
A: There are Steiner triple systems of order $25 \equiv 1 \pmod 6$.  If we take one, and replace each undirected $3$-cycle with a (coherent) directed $3$-cycle, we get a everyone's a pseudo-winner tournament.
A: The answer is a well-known graph generalizing a classical 3-vertex construction:

To see that RPS-25 satisfies the conditions, it's enough to check that each shape beats exactly $12$ others. As a result, if shape $A$ loses to shape $B$, then among the $12$ shapes that shape $A$ beats, at most $11$ can lose to shape $B$: so there must be a shape $C$ that beats $B$ and loses to $A$.
(And so it's easy to find other examples, for any odd $n$: just take any regular tournament.)
