Rock Paper Scissors Lizard Spock, Round Table The question I have is can you order all the elements around a round table such that for each element the members on the left side are inferior and the members on the right side are superior (when seated at the table, facing inward).
I believe the answer is yes. But I want a formal proof for it.
ETA: The final answer is NO, Barry Cipra provided a counterexample with 7 elements.
Example with 3, clockwise order: Paper Rock Scissors
Example with 5: Scissors Lizard Paper Spock Rock
Example with 9: Rock Scissors Spiderman Wizard Lizard Paper Glock Batman Spock
The general case has 2N+1 elements. For each element there are N inferior elements and N superior elements. Any element is equal to itself. Imagine a society where N is huge. For any individual 1/2 of everyone else is superior, and the other 1/2 is inferior. Note that the relationships are symmetric: If A is superior to B then B is inferior to A. Can you always distribute this huge population in a circle such that anyone to the left is inferior and anyone to the right is superior?
ETA: Here are some helpful images:
5 elements: http://www.linuxmotors.com/ScLiPaSpRo.jpg
9 elements: RPS9.jpg
15 elements: http://www.linuxmotors.com/RPS15.jpg
Similiar images are readily available with google image search. Most diagrams have the arrows going willy-nilly (some left, some right). My question is can the elements be reordered such that all the arrows go in the same direction.
Somehow to me it seems like a deep, interesting question, that would have a cool, easy proof.
 A: Here's a counterexample with $2N+1=7$:
Suppose we have the (non-transitive) inequalities 

  
*
  
*Rock < Paper < Scissors < Rock
  

as usual, and 

  
*
  
*Batman < Superman < Aquaman < Batman.
  

Let's interleave these with

  
*
  
*Rock < Batman < Paper, Scissors
  
*Paper < Superman < Scissors, Rock
  
*Scissors < Aquaman < Rock, Paper
  

At this point, Rock, Paper, and Scissors are each inferior to two others and superior to three, while Batman, Superman, and Aquaman are each superior to two others and inferior to three.  
Now add a Wizard with inequalities

  
*
  
*Rock, Paper, Scissors < Wizard < Batman, Superman, Aquaman
  

Now everyone is inferior to three and superior to three.  
Wherever the Wizard sits, Rock, Paper, and Scissors must be to his left (and Batman, Superman, and Aquaman to his right, but it suffices to look at just one side).  If the desired condition were satisfied, then whichever one is seated to Wizard's immediate left would have to be superior to both the other two.  But that's not the case.
A: This depends on what you mean by "inferior" and "superior". It is trivial to arrange things so that $x$ is always "superior" to its predecessor and "inferior" to its successor, going around a circle: just put the elements in a circle, and define inferior = to the left of $x$ and superior = to the right of $x$.
The problem is, this relationship is not transitive: if $a$ is inferior to $b$, and $b$ is inferior to $c$, but it does not follow that $a$ is inferior to $c$. In fact with a circular arrangement, this cannot always hold. Thus I may be able to beat the guy on my left, but who is to say if I can beat the guy on his left?
Such non-transitive relationships have properties that are occasionally useful. After, that is exactly why we have "rock, paper, scissors". If rock always won, it would be pointless. But usually, transitive relationships are much richer.
