The complete graph $K_{14}$ is much bigger than needed for this result; here are three proper subgraphs with the same property.
I. Every $2$-coloring of the complete bipartite graph $K_{3,7}$ contains a monochromatic $C_4$.
Let $K_{3,7}$ have partite sets $V_1,V_2$ with $|V_1|=3$ and $|V_2|=7$. Each vertex in $V_2$ is joined by edges of one color to two vertices in $V_1$. By the pigeonhole principle, two vertices in $V_2$ are joined by edges of the same color to the same two vertices in $V_1$.
II. Every $2$-coloring of the complete graph $K_6$ contains a monochromatic $C_4$. (This was shown in Misha Lavrov's answer; the following argument is perhaps slightly simpler.)
We may assume that there is a vertex $x$ which is incident with at most two blue edges. In fact, we may assume that $x$ is incident with exactly two blue edges; otherwise $x$ would be incident with four red edges, call them $xa_1$, $xa_2$, $xa_3$, $xa_4$, and then we could consider a vertex $y\notin\{x,a_1,a_2,a_3,a_4\}$ and proceed as in paw88789's answer. (It may be even easier to observe that the subgraph induced by $\{a_1,a_2,a_3,a_4\}$ contains either a red $P_3$ or a blue $C_4$.)
So let $x$ be incident with exactly two blue edges, $xy$ and $xz$. If one of the remaining three vertices is joined by blue to both $y$ and $z$, then we have a blue $C_4$. On the other hand, if each of those three vertices is joined by red to a vertex in $\{y,z\}$, then two of them are joined by red to the same vertex in $\{y,z\}$, and also to $x$, making a red $C_4$.
III. Every $2$-coloring of the complete bipartite graph $K_{5,5}$ contains a monochromatic $C_4$.
The proof is left as an exercise for the reader.
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.
Best 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.