Four (50 points) Let $k \in N^{*}$, if it is possible to color all positive integers with $k$ colors, and there exists a function $f: Z^{+} \rightarrow Z^{+}$ satisfying:
(1) For any same-colored positive integers $m, n$ (which can be the same), $f(m+n)=f(m)+f(n)$;
(2) There exist $m, n \in N^{*}$ (which can be the same) such that $f(m+n) \neq f(m)+f(n)$. Find the minimum value of $k$.


Four, Solution: The minimum value of $k$ is 3.
First, we provide a construction for $k=3$. Let $f(n)=\left\{\begin{array}{ll}2 n, & 3 \mid n \\ n, & 3 \nmid n\end{array}\right.$, then $f(1)+f(2)=3 \neq f(3)$, satisfying condition (2). If we color the numbers that are congruent to $0, 1, 2 \pmod{3}$ with three different colors, then:
- If $x \equiv y \equiv 0 \pmod{3}$, then $x+y \equiv 0 \pmod{3}$, and $f(x+y)=2(x+y)=f(x)+f(y)$;
- If $x \equiv y \equiv 1 \pmod{3}$, then $x+y \equiv 1$ or $2 \pmod{3}$, and $f(x+y)=x+y=f(x)+f(y)$;
- If $x \equiv y \equiv 2 \pmod{3}$, then $x+y \equiv 1$ or $2 \pmod{3}$, and $f(x+y)=x+y=f(x)+f(y)$;
All of these satisfy condition (1), so $k=3$ satisfies the problem.

Next, we prove that $k=2$ does not satisfy the problem. Suppose otherwise, for a function $f$ and coloring scheme that satisfies (1), setting $m=n$ in (1) gives $f(2 n)=2 f(n), \forall n \in N_{+}$(1); for $\forall n \in N_{+}$, by (1), $f(2 n)=2 f(n), f(4 n)=4 f(n)$, $f(6 n)=2 f(3 n)$, we consider the following cases:
1. If $n$ and $2 n$ are the same color, then $f(3 n)=f(2 n)+f(n)=3 f(n)$;
2. If $2 n$ and $4 n$ are the same color, then $f(3 n)=\frac{1}{2} f(6 n)=\frac{1}{2}(f(4 n)+f(2 n))=3 f(n)$;
3. If $2 n$ is a different color from both $n$ and $4 n$, then $n$ and $4 n$ are the same color. In this case, if $n$ and $3 n$ are the same color, then
   $$
   f(3 n)=f(4 n)-f(n)=3 f(n);
   $$
   If $n$ and $3 n$ are different colors, then $2 n$ and $3 n$ are the same color, so
   $$
   f(3 n)=f(4 n)+f(n)-f(2 n)=3 f(n).
   $$

Thus, for any positive integer $n$, we always have $f(3 n)=3 f(n)$.
We now prove that $f(n)=n f(1)$ (i.e., satisfying (1) does not satisfy (2)).
If not, there exists a positive integer $m$ such that $f(m) \neq m f(1)$. We choose $m$ to be the smallest positive integer satisfying $f(m) \neq m f(1)$, then $m \geq 5$ and is odd (otherwise, if $m$ is even, by the minimality of $m$, we have $f\left(\frac{m}{2}\right)=\frac{m}{2} f(1)$, thus $f(m)=2 f\left(\frac{m}{2}\right)=m f(1)$, a contradiction). Since $m \geq 5$, we have $\frac{m-3}{2}<\frac{m+3}{2}<m$, and by the minimality of $m$, we have $f\left(\frac{m-3}{2}\right)=\frac{m-3}{2} \cdot f(1), f\left(\frac{m+3}{2}\right)=\frac{m+3}{2} \cdot f(1)$. If $\frac{m+3}{2}$ and $\frac{m-3}{2}$ are the same color, then $f(m)=f\left(\frac{m+3}{2}\right)+f\left(\frac{m-3}{2}\right)=m f(1)$, a contradiction, so $\frac{m+3}{2}$ and $\frac{m-3}{2}$ are different colors, thus $m$ is exactly the same color as one of $\frac{m+3}{2}, \frac{m-3}{2}$. We consider these two cases:
1. If $m$ and $\frac{m-3}{2}$ are the same color, then $f(m)+f\left(\frac{m-3}{2}\right)=f\left(3 \cdot \frac{m-1}{2}\right)=3 f\left(\frac{m-1}{2}\right)=\frac{3(m-1)}{2} \cdot f(1)$, so $f(m)=m f(1)$, a contradiction;
2. If $m$ and $\frac{m+3}{2}$ are the same color, then $f(m)+f\left(\frac{m+3}{2}\right)=f\left(3 \cdot \frac{m+1}{2}\right)=3 f\left(\frac{m+1}{2}\right)=\frac{3(m+1)}{2} \cdot f(1)$, so $f(m)=m f(1)$, a contradiction.
In summary, $k=2$ does not satisfy the problem, so the minimum value of $k$ is 3.