Problem: Find the smallest positive integer $k$ for which
there exists a coloring of the positive integers
$\ZZ_{>0}$ with $k$ colors and a function $f \colon \ZZ_{>0} \to \ZZ_{>0}$
with the following two properties:
\begin{enumerate}[(i)]
  \ii For all positive integers $m$, $n$ of the same color,
  $f(m+n)=f(m)+f(n)$.
  \ii There are positive integers $m$, $n$
  such that $f(m+n) \neq f(m)+f(n)$.
\end{enumerate}

Solution: Answer: $k=3$.

Construction for $k=3$:
let
\[ f(n) =
  \begin{cases} n/3 & n\equiv 0 \pmod 3 \\
  n & \text{else} \end{cases} \]
and color the integers modulo $3$.

Now we prove that for $k=2$ a function $f$ obeying (i) must be linear,
even if $f \colon \ZZ_{>0} \to \RR_{>0}$.
Call the colors blue/red and WLOG $f(1) = 1$.

First, we obviously have:
\begin{claim*}
  $f(2n) = 2f(n)$ for every $n$.
\end{claim*}

Now we proceed by induction in the following way.
Assume that $f(1) = 1$, $f(2) = 2$, \dots, $f(2n) = 2n$.
For brevity let $m = 2n+1$ be red
and assume for contradiction that $f(m) \neq m$.

The proof now proceeds in four steps.
First:
\begin{itemize}
  \ii The number $m-2$ must be blue.
  Indeed if $m-2$ was red we would have $f(2m-2) = f(m) + f(m-2)$
  which is a contradiction as $f(2m-2) = 2f(m-1) = 2m-2$ and $f(m-2)=m-2$.

  \ii The number $2$ must be red.
  Indeed if it was blue then $f(m) = f(2) + f(m-2) = m$ contradiction.
\end{itemize}
Observe then that $f(m+2) = f(m)+2$ since $m$ and $2$ are both red.
Now we consider two cases:
\begin{itemize}
  \ii If $m+2$ is red, then
  $f(2m+2) = f(m+2) + f(m) = (f(m) + 2) + f(m)$.
  But $f(2m+2) = f(4n+4) = 4f(n+1) = 2m+2$, contradiction.

  \ii If $m+2$ is blue,
  then $2f(m) = f(2m) = f(m+2) + f(m-2) = f(m) + 2 + (m-2)$.
  So then $f(m) = m$ again a contradiction.
\end{itemize}

So $f(m) = m$, which completes the induction.