Generalizing Rock, Paper, Scissors game? Problem of playing rock-paper-scissors with $n$ elements and keeping it balanced

How to extend the rock-paper-scissors game to more than just $3$
  elemetns, and keep it balanced?
In a balanced game, optimal strategy would be to pick every choice
  with equal probability.

Is there a general rule that can be used to efficiently evaluate a winner between any two distinct options?
For example, there is one generalization to five elements, rock-paper-scissors-lizard-spock:

 A: Stating the problem
Let $G(n)$ be a rock-paper-scissors game with $n\in\mathbb N$ elements.
Let those $n$ elements be the fist $n$ nonnegative integers $\{0,1,2,\dots,n-1\}=N_n$. 
For every $n_0,n_1\in N$, if $n_0$ beats $n_1$, then $n_0$ is a "counter" to $n_1$, and $n_1$ is a "anticounter" to $n_0$. Counters and anticounters of $n_0$ are together called "interactions" of $n_0$.
We want to establish a rule for counters and anticounters of every $n_0\in N_n$, such that:


*

*$n_0$ has an equal number of counters and anticounters - interactions are balanced. 

*$n_0$'s interactions in $G(n)$ should be kept in $G(n+1)$.
Solution to the problem
The first property requires $n=2k+1,k\in\mathbb N$ to be odd.
If we start observing first few cases of $G(n)$, you'll notice that to satisfy the second property, $n_0=m$ needs to be a counter to $m+1,m-2,m+3,m-4,\dots$
This is finally equivalent to the following rule: The winner between $n_0,n_1\in N_n$ is:


*

*if $n_0,n_1$ have the same parity (both odd or both even), then winner is: $\max\{n_0,n_1\}$

*else, if $n_0,n_1$ have $\ne$ parity, then the winner is: $\min\{n_0,n_1\}$
This rule works for defining counters and anticounters for every element.
This rule allows a balanced rock-paper-scissors to be played with any number of odd elements, and to easily add/remove interactions to extend or reduce the game by $2$ elements.
For example, $n=3,5$ cases and with the classic labels to elements $0,1,2,3,4$ :


Generalization to countable sets
Notice that the general observed rule does not depend on $n$. 
Thus if we let $n\to\infty$, we can play rock-paper-scissors on $\mathbb N_0 = \mathbb N \cup \{0\}$.
We can also include negative numbers and extend it to the whole integers $\mathbb Z$, as the "parity" and "$\lt$" are still defined on integers.
We can also play it on any countable set. We establish a bijection from some countable set $X$, such as rationals $X=\mathbb Q$ for example, to either $\mathbb N_0$ or $\mathbb Z$, and call it $B(x)$.
Then the winner between $x_1,x_2\in X$ is:
$$=\begin{cases}
  \displaystyle \max\{B(x_1), B(x_2)\}, & B(x_1)\bmod2=B(x_2)\bmod2\\
  \displaystyle \min\{B(x_1), B(x_2)\}, & B(x_1)\bmod2\ne B(x_2)\bmod2
\end{cases}$$
I believe this rule is the most "balanced" way to extend rock-paper-scissors to any finite and countable sets.
Additionally, by symmetry, we can invert the rule and keep all the properties. That is, invert the conditions for min and max calculations.
A: For any odd number $n$, consider a game with outcomes in the range $0$ to $n-1$. Outcome $x$ beats outcome $y$ iff $(y-x) \bmod{n} < \frac{n}{2}$. In other words, each outcome beats the next half of the outcomes, and loses to the other half.
A: This answer is meant as an addition to the answers of Vepir and Thomas.
I was wondering in how many ways we could generalize rock-paper-scissors. Here my findings.
We can represent the rules of the game with a directed graph like the one in the question. An arc from A to B means A beats B. For every pair of vertices A and B we need to have either A → B or B → A. Such a graph is called a tournament. To keep the game balanced every vertex needs to have the same indegree as outdegree. A tournament with that property is called a regular tournament. Note that a regular tournament needs to have an odd number of vertices.
In the table below we can see the number of (nonisomorphic) regular tournaments of a given vertex count.
We can introduce extra requirements to prefer one regular tournament over an other:

*

*We can require the graph to be vertex-transitive. This adds a nice symmetry so that, loosely speaking, every vertex is the same.
Note that a vertex-transitive tournament is automatically regular.

*We can require that we have to be able to build the tournament by starting at $n=1$, and then adding two vertices and multiple arcs at each step, while at each step having a regular tournament. This is similar to Vepir's "$n_0$'s interactions in $G(n)$ should be kept in $G(n+1)$". I'll call this propery extended.

*We can combine these two properties: We start at $n=1$, add two vertices and multiple arcs at each step, while at each step having a vertex-transitive tournament. I'll call these tournaments extended vertex-transitive tournaments.

Counting all the possibilities, we get:
\begin{array}{|l|c|c|c|c|c|c|c|}
\hline
& 1 & 3 & 5 & 7 & 9 & 11 & 13 \\ \hline
\text{Regular tournaments (OEIS A096368)} & 1 & 1 & 1 & 3 & 15 & 1223 & 1495297 \\ \hline
\text{Vertex-transitive tournaments (OEIS A346179)} & 1 & 1 & 1 & 2 & 3 & 4 & 6 \\ \hline
\text{Extended regular tournaments} & 1 & 1 & 1 & 2 & 8 & 350 & ? \\ \hline
\text{Extended vertex-transitive tournaments} & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline
\end{array}
Sequences: OEIS A096368, OEIS A346179
It seems that there is only one way (up to isomorphism) to generalize rock-paper-scissers in a way that it is extended vertex-transitive.
I tested the formulas of Vepir and Thomas and it turns out they both generate exactly this graph!
