We write $a \sim b$ iff we can prove that $a$ and $b$ must have the same color. This clearly is an equivalence relation. Observation 1: if $a \equiv b \mod k$, then $a \sim b$. Proof: W.l.o.g. $a<b$ and write $b=a+mk$, $m>0$. Then we have $b \sim |b-k| = b-k \sim |b-k-k| = b-2k \sim ... \sim |b-mk| = b-mk =a$. Observation 2: if $a \equiv -b \not \equiv 0 \mod k$, then $a \sim b$. Proof: by observation 1, we may assume $0<a<k$ ($a=k$ is excluded by $a \not \equiv 0 \mod k$). Then $a \sim |a-k| = k-a \sim b$, the latter by observation $1$ again. Observation 3: if $b \equiv a-n \not \equiv 0 \mod k$ then $a \sim b$. Proof: we have $a \sim n-a \sim b$, the latter by observation 2. We finish the problem by remembering that $\gcd(n,k)=1$, thus $-n, -2n , -3n , ... -(k-1)n , -kn \mod k$ are all residue classes. And if $a_1,...,a_k$ from the allowed interval represent them in that order, we always have $a_i \sim a_{i+1}$ by the last observation. Thus all have the same color.

I think it should be ZetaX wrote: Observation 2: if $a \equiv -b \equiv 0 \mod k$, then $a \sim b$. Observation 3: if $b \equiv a-n \equiv 0 \mod k$ then $a \sim b$. Proof: we have $a \sim n-a \sim b$, the latter by observation 2

And why would you think that¿ This special cases would be useless.