graph theory
In complete graph at 2017 vertices, all the edges were painted in 10 colors. Prove, that there is a cycle of odd length, all the edges of which are painted in the same color.
I have only one idea: that this graph has at least $\left[\frac{2016\cdot 2017}{2\cdot 10} + 1\right]$ edges of the same color.
As the question correctly observes, there are ten odd-degree nodes that need to be revisited as part of a cyclic tour of the edges. If you double some of the edges (making a multigraph without loops), you can reduce the number of nodes with odd degree. Then for an Eulerian path on this generated multigraph, you'd need to find the least number of edges you can add to make all but two of the nodes of even degree. For an Eulerian cycle as required, you need to eliminate all odd nodes by adding such edges.
There's clearly a solution for $7$ added edges, as you say, illustrated below, and the $10$ odd nodes require at least $5$ added edges to resolve to even degree in a multigraph, so demonstrating that neither $5$ nor $6$ added edges are sufficient is required. The odd-degree nodes are here organized into four sets on the edges of the grid. If four adjacent odd nodes are connected with a doubled edge, choosing one from each set of adjacent blocks, the remaining two nodes (eg. $(0,3)$ and $(3,3)$) require 3 additional edges to connect, as per the illustration. If only three adjacent nodes are connected then both of the remaining pairs require at least two edges each to connect. In both cases $7$ doubled edges is the minimum, leading to the minimum tour length of $31+7=38$ edge traverses.
A question in which the adjacency of odd nodes was less straightforward, or in which edges are different lengths, could be significantly harder to answer.
In fact, if the edges of $K_n$ are colored with at least $n$ different colors, then there is a rainbow triangle (a triangle whose edges are all of different colors).
Proof by induction of $n$. The result is trivial for $n\le3$. Suppose $n\gt3$. Choose a vertex $u$ and consider the graph $K_n-u=K_{n-1}$.
If $K_{n-1}$ has edges of at least $n-1$ different colors, then it contains a rainbow triangle by the induction hypothesis.
On the other hand, if $K_{n-1}$ has edges of at most $n-2$ different colors, then among the colors of the edges incident with $u$ there must be at least two colors which do not occur in $K_{n-1}$. If the edges $uv$ and $uw$ have two different colors not occurring in $K_{n-1}$, then $uvw$ is a rainbow triangle.
P.S. Here's an alternative presentation, really the same argument. Call the vertices $v_1,v_2,\dots,v_n$. Let's say that a color (call it "red") occurs at a vertex $v_i$ if there is a red edge joining $v_i$ to an earlier vertex, i.e., a red edge $v_jv_i$ for some $j\lt i$. Let $N_i$ be the number of new colors occurring at $v_i$, i.e., colors which occur at $v_i$ but do not occur at any $v_j$ where $j\lt i$.
Then $N_1+N_2+\cdots+N_n\ge n$, since there are edges of at least $n$ different colors. Since $N_1=0$, we must have $N_i\gt1$ for some $i$. Finally, if two new colors occur at $v_i$, then $v_i$ is in a rainbow triangle.
Best Answer
It is a well-known theorem that a graph is bipartite if and only if it contains no odd cycles. So, you are essentially being asked to show that $K_{2017}$ cannot be expressed as the union of $10$ edge-disjoint bipartite graphs.
Let $G$ be a graph on 2017 vertices that is the union of ten edge-disjoint bipartite graphs. Since $2017>2^{10}$, by the pigeonhole principle there must be two vertices that are in the same part of all ten of those bipartitions. Since no vertex is connected to a vertex in the same part of a bipartite graph, those two vertices cannot be adjacent in $G$. Therefore $G$ cannot be complete.