WLOG let $1$ be red.
$\textbf{Case 1: }$ $2$ is red.
For any red integer $n$ we have $n+1$ will also be red. Thus, all integers are red.
$\textbf{Case 2: }$ $2$ is green and $3$ is red.
For any green integer $n$ we have $n+2$ will also be green. Thus, all even integers are green.
We have $3+1+1=5$ will be red. From here, we can add $1+1$ to the LHS and $2$ to the RHS to obtain the next odd integer. Thus, all odd integers greater than $1$ are red.
$\textbf{Case 3: }$ $2$ and $3$ are green.
As before, we know all even integers are green.
For any green integer $n$, we have $n+2$ will also be green. Since $3$ is green, it follows that all odd integers greater than $3$ are also green.
In summary, we can have:
$\textbf{1. }$ All integers are red
$\textbf{2. }$ All odd integers are red and all even integers are green
$\textbf{3. } 1$ is red and all other integers are green
Plus the cases where the colors are swapped