Find the smallest positive integer $k$ for which there exists a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f:\mathbb{Z}_{>0}\to \mathbb{Z}_{>0}$ 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 $f(m+n)\ne f(m)+f(n).$

[i]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 distinct.[/i]
1. **Define the problem and the goal**: We need to find the smallest positive integer \( k \) such that there exists a coloring of the positive integers \(\mathbb{Z}_{>0}\) with \( k \) colors and a function \( f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0} \) satisfying:
   - (i) For all positive integers \( m, n \) of the same color, \( f(m+n) = f(m) + f(n) \).
   - (ii) There exist positive integers \( m, n \) such that \( f(m+n) \neq f(m) + f(n) \).

2. **Show that \( k = 3 \) works**:
   - Consider the coloring with three colors: red, blue, and green.
   - Color all \( 0 \pmod{3} \) integers red, all \( 1 \pmod{3} \) integers blue, and all \( 2 \pmod{3} \) integers green.
   - Define the function \( f \) as follows:
     \[
     f(x) = \begin{cases} 
     x & \text{if } x \equiv 0 \pmod{3}, \\
     2x & \text{otherwise}.
     \end{cases}
     \]

3. **Verify condition (i)**:
   - For \( m, n \) both red (\( m, n \equiv 0 \pmod{3} \)):
     \[
     f(m) + f(n) = m + n = f(m+n).
     \]
   - For \( m, n \) both blue (\( m, n \equiv 1 \pmod{3} \)):
     \[
     f(m) + f(n) = 2m + 2n = 2(m+n) = f(m+n).
     \]
   - For \( m, n \) both green (\( m, n \equiv 2 \pmod{3} \)):
     \[
     f(m) + f(n) = 2m + 2n = 2(m+n) = f(m+n).
     \]

4. **Verify condition (ii)**:
   - Consider \( m = 1 \) and \( n = 2 \):
     \[
     f(1) = 2, \quad f(2) = 4, \quad f(3) = 3.
     \]
     \[
     f(1) + f(2) = 2 + 4 = 6 \neq 3 = f(3).
     \]
   - Thus, \( f(1+2) \neq f(1) + f(2) \), satisfying condition (ii).

5. **Show that \( k = 2 \) fails**:
   - Assume \( k = 2 \) works. Let the two colors be red and blue.
   - Let \( f(1) = c \). We will prove by strong induction that \( f(x) = cx \) for all \( x \geq 1 \).
   - Base case: \( f(1) = c \).
   - Inductive step: Assume \( f(i) = ci \) for all \( 1 \leq i \leq k \). Consider \( f(k+1) \):
     - If \( k+1 \) is even, \( k+1 = 2m \):
       \[
       f(k+1) = f(m) + f(m) = cm + cm = c(k+1).
       \]
     - If \( k+1 \) is odd, \( k+1 = 2m+1 \):
       \[
       f(k+1) = f(1) + f(2m) = c + 2cm = c(2m+1) = c(k+1).
       \]
   - Thus, \( f(x) = cx \) for all \( x \geq 1 \), contradicting condition (ii).

6. **Show that \( k = 1 \) fails**:
   - If \( k = 1 \), all integers are the same color, making conditions (i) and (ii) mutually exclusive.

Conclusion:
- \( k = 1 \) and \( k = 2 \) do not work.
- \( k = 3 \) works.

The final answer is \( \boxed{ k = 3 } \).