2. Let $k$ be a positive integer. Suppose that all positive integers can be colored using $k$ colors, and there exists a function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying:
(1) For any positive integers $m, n$ of the same color (allowing $m = n$), we have $f(m+n)=f(m)+f(n)$;
(2) There exist positive integers $m, n$ (allowing $m = n$) such that $f(m+n) \neq f(m)+f(n)$.
Find the minimum value of $k$.


2. The minimum value of $k$ is 3.
First, construct an example for $k=3$.
Let $f(n)=\left\{\begin{array}{ll}2 n, & n \equiv 0(\bmod 3) ; \\ n, & n \equiv 1,2(\bmod 3)\end{array}\right.$
Then $f(1)+f(2)=3 \neq f(3)$, satisfying condition (2).
At the same time, color the numbers that are congruent to $0, 1, 2 \pmod{3}$ with three different colors, respectively. Thus,
(i) For any $x \equiv y \equiv 0(\bmod 3)$, we have
$x+y \equiv 0(\bmod 3)$
$\Rightarrow f(x+y)=\frac{x+y}{3}=f(x)+f(y)$;
(ii) For any $x \equiv y \equiv 1(\bmod 3)$, we have
$x+y \equiv 2(\bmod 3)$
$\Rightarrow f(x+y)=x+y=f(x)+f(y)$;
(iii) For any $x \equiv y \equiv 2(\bmod 3)$, we have
$x+y \equiv 1(\bmod 3)$
$\Rightarrow f(x+y)=x+y=f(x)+f(y)$.
Thus, condition (1) is also satisfied.
Therefore, $k=3$ meets the requirements.
Next, prove that $k=2$ does not hold.
It suffices to prove that for any function $f$ and coloring scheme satisfying condition (1) when $k=2$, we have
$f(n)=n f(1)$ (for any $n \in \mathbf{Z}_{+}$),
which contradicts condition (2).
In condition (1), take $m=n$, then
$f(2 n)=2 f(n)$ (for any $n \in \mathbf{Z}_{+}$).
Next, prove:
$f(3 n)=3 f(n)$ (for any $n \in \mathbf{Z}_{+}$).
For any positive integer $n$, by equation (2) we know
$f(2 n)=2 f(n), f(4 n)=4 f(n)$,
$f(6 n)=2 f(3 n)$.
If $n$ and $2 n$ are the same color, then
$f(3 n)=f(2 n)+f(n)=3 f(n)$,
equation (3) holds;
If $2 n$ and $4 n$ are the same color, then
$$
\begin{array}{l}
f(3 n)=\frac{1}{2} f(6 n)=\frac{1}{2}(f(4 n)+f(2 n)) \\
=3 f(n),
\end{array}
$$
equation (3) also holds.
Otherwise, $2 n$ is a different color from both $n$ and $4 n$, so $n$ and $4 n$ are the same color. In this case, if $n$ and $3 n$ are the same color, then
$$
f(3 n)=f(4 n)-f(n)=3 f(n),
$$
equation (3) holds;
If $n$ and $3 n$ are different colors, then $2 n$ and $3 n$ are the same color,
$$
f(3 n)=f(4 n)+f(n)-f(2 n)=3 f(n) \text {, }
$$
equation (3) also holds.
Thus, equation (3) is proven.
Assume the proposition (1) does not hold. Then there exists a positive integer $m$ such that $f(m) \neq m f(1)$.
Without loss of generality, take the smallest $m$, then by equations (2) and (3), $m \geqslant 5$, and $m$ is odd. Otherwise, by the minimality of $m$,
$f\left(\frac{m}{2}\right)=\frac{m}{2} f(1)$.
Thus, $f(m)=2 f\left(\frac{m}{2}\right)=m f(1)$, a contradiction.
Consider $\frac{m-3}{2}<\frac{m+3}{2}<m$ these three numbers.
Similarly, by the minimality of $m$,
$f\left(\frac{m-3}{2}\right)=\frac{m-3}{2} f(1)$,
$f\left(\frac{m+3}{2}\right)=\frac{m+3}{2} f(1)$.
Thus, $\frac{m-3}{2}$ and $\frac{m+3}{2}$ are different colors. Otherwise,
$f(m)=f\left(\frac{m-3}{2}\right)+f\left(\frac{m+3}{2}\right)=m f(1)$,
a contradiction.
Therefore, $m$ is exactly the same color as one of $\frac{m-3}{2}, \frac{m+3}{2}$.
Let $m$ be the same color as $\frac{m+3 p}{2}(p \in\{-1,1\})$.
Notice that, $\frac{m+p}{2}<m$.
Then $f(m)+f\left(\frac{m+3 p}{2}\right)=f\left(3 \times \frac{m+p}{2}\right)$
$=3 f\left(\frac{m+p}{2}\right)=\frac{3(m+p)}{2} f(1)$
$\Rightarrow f(m)=m f(1)$,
a contradiction.
Thus, proposition (1) is proven, i.e., the minimum value of $k$ is 3.