A hard graph theory problem on friends and enemies Well, I guess it's to be solved using graph theory, but how? Any hints welcome. 
The problem goes as follows: we have a group $n\ge 9$ people. Two people can either be enemies, friends or be neutral to each other. There are two conditions: every person has the same number of enemies and friends and there is a person, that is either a friend or an enemy of at least six people (for example, he can have three friends and three enemies). 
What we need to prove is that we can make at least two people (but not specifically two) neutral to each other, but not all of them, so that in the end every person will still have the same number of enemies and friends
Any hints welcome. I guess we could try induction on $n$...? The base case seems to work, but I think its rather to be approached using graph theory. Maybe Hall's theorem?
Clarification: by "every person has the same number of enemies and friends", I mean a person $p$ must have $n$ friends and $n$ enemies and another person $p'$ has $m$ friends and $m$ enemies, but we can have $n\neq m$. And by "if two or more, but not all people become neutral", I mean that some subset of at least $2$ people are made pairwise neutral, as in every pair of people in that subset are made neutral.
 A: Suppose integer $n\ge 9$, and adopt the edge coloring scheme on $K_n$ proposed by @JohnWatson: edges between friends are green, between enemies red, and all other edges are white.
We are given these restrictions:


*

*At each node the number of red edges is the number of green edges.  

*Some node has at least six edges that are red or green.  

*Some edge is white (neither red nor green).  


Given the somewhat garbled phrasing of the Question, it seems likely to me that a further restriction may have been originally intended:


*

*No node has only white edges, so each node has at least one red and one green edge.


In any case we will show that some edge coloring exists to satisfy these restrictions.  The case where $n$ is even is somewhat easier, because we can appeal to the existence of a $1$-factorization of $K_n$, a partition of the edges into $n-1$ parallel classes each of which is a perfect matching or 1-factor.  Construction of such a $1$-factorization for even $n$ amounts to a round-robin tournament, scheduling of which has been variously described in previous Questions, e,g, here, here, and here.
So for even $n\gt 9$ we choose six of these disjoint matchings and color three of them green and the other three red (leaving the rest of the edges white).  Thus every node has exactly three green edges and three red edges, and the remaining edges on any node (since $n-1 \gt 6$) are white.
For the cases where $n\ge 9$ is odd, there is no perfect matching (since it takes two nodes to make an edge).  One simple idea is to "remove" a node and color the remaining complete graph on $n-1$ nodes subject to the above restrictions, but I added the final restriction (that no node has only white edges) in part to block this approach (the "removed" node would have no green or red edges).  Instead we present a construction involving Hamiltonian circuits.
When $n$ is odd, Walecki's Theorem (1890s) says $K_n$ has an edge partition into $(n-1)/2$ Hamiltonian circuits, each a cycle of length $n$.  If $n\ge 11$, then we can choose two Hamiltonian circuits to color green and two to color red, leaving one or more Hamiltonian circuits colored white.  Since such a circuit contributes two edges for each node, all the nodes have eight edges that are green or red (and there are some white edges left over).
That leaves just the "special case" $n=9$, for which we can give a particular solution.   Let's start with seven nodes and draw $K_7$ as a regular heptagon with two seven-pointed stars, as taken from Wikipedia:

This figure illustrates the Hamiltonian decomposition of $K_7$ described above, using a different color for each of the $(7-1)/2 = 3$ Hamiltonian circuits.  Keeping the red and green circuits (and switching the blue one to white to "suppress" it), we have only to attach two more nodes, each with one red and one green edge to connect to a (shared) pair of the earlier nodes.
The resulting graph on nine nodes will have the latter pair of nodes with three red and three green edges, and the other conditions are also met.
