Consider a fixed vertex $v$, and suppose $n-1=kq+r$ with $0 \le r <k$ (that is, $r$ is the remainder on dividing $n-1$ by $k$, and $q$ is the quotient.
Then there are $k$ at distance 1 from $v$, another $k$ at distance 2, and so on, up to a final $k$ at distance $q$ from $v$, and the remaining (if any) $r$ are at distance $q+1$ from $v$. This gives the total contribution from $v$ as
$$k(1+2+...+q)+r(q+1)=k\cdot q(q+1)/2 +r(q+1).$$
Since there are $n$ vertices the above must be multiplied by $n$ for the final sum.
EDIT: perhaps this answer has to be divided by 2, since it counts each distance between say $v$ and $w$ twice.
I have the answer for c = 1/2.
Just consider the white edges. As you might know, for every (undirected) graph you can can delete at most half of edges to get a bipartite graph. We'll call the partitions A and B. So there are at least half of the white edges between the group A and B. So if we color all the vertices of which ever group to white, there are at least half of the white edges which is connected to at least one white vertex. So we can choose to paint all the vertices of A to white, or all the vertices of B to white. (We'll choose it later)
Now that we partitioned the vertices into two groups, we want to see which partition should we paint to black, the other partition would be colored to white, and as mentioned, for whites the conditions is held.
The black edges between two partitions, will be connected to at least one black vertex whichever way. Now see the black edges inside the partitions. If black edges in partition A is more, we'll color all the vertices of the A to black. And if black edges in partition B is more, we'll do otherwise.
You can see for white and black the conditions is held.
Proof for the theorem used in this solution: Given a graph, partition it into 2 groups whichever way. If there was a vertex in one group, which more than half of its neighbors are in its group, move the vertex to the other partition. You can see this operation will end, and when ended, for each vertex at least half of its neighbors is in the other group. So at least half of the edges is in-between the partitions.
Update: In this proof, by neighbors we mean the number of edges connected to one vertex.
Best Answer
consider the set of all pairwise distance of black, if distance $j$ is not in it, then $x \mapsto x+j$ is a bijection between black and white.
If we take multiplicity into consideration, the multiplicity of black $j$ is number of points that itself is black and maps to black, it is the same as # of points that is white and maps to white.