Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$.

In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.

Merlijn Staps, the Netherlands
The answer is $k=3$.
First we show that there is such a function and coloring for $k=3$. Consider $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ given by $f(n)=n$ for all $n \equiv 1$ or 2 modulo 3, and $f(n)=2 n$ for $n \equiv 0$ modulo 3. Moreover, give a positive integer $n$ the $i$-th color if $n \equiv i$ (3).
By construction we have $f(1+2)=6 \neq 3=f(1)+f(2)$ and hence $f$ has property (ii).
Now let $n, m$ be positive integers with the same color $i$. If $i=0$, then $n+m$ has color 0, so $f(n+m)=$ $2(n+m)=2 n+2 m=f(n)+f(m)$. If $i=1$, then $n+m$ has color 2, so $f(n+m)=n+m=f(n)+f(m)$. Finally, if $i=2$, then $n+m$ has color 1, so $f(n+m)=n+m=f(n)+f(m)$. Therefore $f$ also satisfies condition (i).

Next we show that there is no such function and coloring for $k=2$.
Consider any coloring of $\mathbb{Z}_{>0}$ with 2 colors and any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Then there exist positive integers $m$ and $n$ such that $f(m+n) \neq f(m)+f(n)$. Choose $m$ and $n$ such that their sum is minimal among all such $m, n$ and define $a=m+n$. Then in particular for every $b<\frac{a}{2}$, we have $f(2 b)=2 f(b)$. Since $a$ is odd, we have $\frac{a+1}{2}<a$ and therefore $f(a+1)=2 f\left(\frac{a+1}{2}\right)=$ $(a+1) f(1)$. As $a-1$ is an even integer smaller than $a$, we have by Lemma 2 that $a-1$ also has the same color as $a$. Hence $2 f(a)=f(2 a)=f(a+1)+f(a-1)=(a+1) f(1)+(a-1) f(1)=2 a f(1)$, which implies that $f(a)=a f(1)$, a contradiction. So $a$ and $a+1$ have different colors.
Since $a-2$ is an odd integer smaller than $a$, by Lemma 1 it has a color different from that of $a$, so $a-2$ and $a+1$ have the same color. Also, we have seen by Lemma 2 that $a-1$ and $a$ have the same color. So $f(a)+f(a-1)=f(2 a-1)=f(a+1)+f(a-2)=(a+1) f(1)+(a-2) f(1)=(2 a-1) f(1)$, from which it follows that $f(a)=(2 a-1) f(1)-f(a-1)=(2 a-1) f(1)-(a-1) f(1)=a f(1)$, which contradicts our choice of $a$ and finishes the proof.