I read this sentence in one combinatorics book.
In graph $K_{100}$, there is a possible way to assigns number (value) from $\{+1,-1\}$ to each edge, so that the sum of all edge values connected to each vertex becomes $+1$.
I try to understand it, but I could not verify it. Please help me.
Proof by induction:
For every complete graph $K_{2n}$ with $2n$ vertices, there is a labeling of the edges with ${-1,+1}$ such that every vertex has $n$ edges labeled $+1$ and $n-1$ edges labeled $-1$.
The claim is trivially true for the null graph since there are no vertices. For the graph $K_2$, label the single edge $+1$.
Take a graph $K_{2n}$ labeled as described above. Divide its set of vertices into two subsets of equal size: $V = V_1 \cup V_2$, $|V_1|=|V_2|=n$. Add two new vertices $x$ and $y$. Add edges and labels as follows:
Edges connecting $x$ to all vertices in $V_1$ are labeled $+1$.
Edges connecting $y$ to all vertices in $V_1$ are labeled $-1$.
Edges connecting $x$ to all vertices in $V_2$ are labeled $-1$.
Edges connecting $y$ to all vertices in $V_2$ are labeled $+1$.
The edge $xy$ is labeled $+1$.
Vertices in $V_1$ each gained one edge labeled $+1$ to $x$ and one labeled $-1$ to $y$, so they now have $n+1$ edges labeled $+1$ and $n$ edges labeled $-1$. Similarly, vertices in $V_2$ each gained two edges of opposite sign, so they also have $n+1$ labeled $+1$ and $n$ labeled $-1$. $x$ has $n$ edges labeled $+1$ to $V_1$, $n$ edges labeled $-1$ to $V_2$, and one edge labeled $+1$ to $y$, for a total of $n+1$ labeled $+1$ and $n$ labeled $-1$. The same way, $y$ also has $n+1$ edges labeled $+1$ and $n$ labeled $-1$.
So we have constructed a labeling of $K_{2n+2}$ that satisfies the induction hypothesis. Such a labeling is possible for all complete graphs with even vertex count.