25. Let $f$ be defined on the set of positive integers such that for all $m, n \in \mathbf{N}^{+}$, we have $f(m \cdot n)=f(m)+f(n)$, and $f(n+1) \geqslant f(n)$. Try to find $f(n)$.


25. Let $r, s, n \in \mathbf{N}$ and $n \geqslant 1, s \geqslant r \geqslant 2, m \in \mathbf{N}^{+}$ satisfy $r^{m} \leqslant s^{m} \leqslant r^{m+1}$ (by the definition of the exponential function, such an $m$ must exist). Taking the logarithm of the inequality, we get
$$
\begin{array}{l}
m \ln r \leqslant n \ln s \leqslant (m+1) \ln r, \\
\frac{m}{n} \leqslant \frac{\ln s}{\ln r} \leqslant \frac{m}{n} + \frac{1}{n}, \\
f\left(r^{\prime \prime \prime}\right) \leqslant f\left(s^{n}\right) \leqslant f\left(r^{m+1}\right), \\
f\left(r^{n}\right) = m f(r) \\
f\left(s^{n}\right) = n f(s),
\end{array}
$$

Since $f$ is non-decreasing, we have
By the conditions,

Thus,
$$
m f(r) \leqslant n f(s) \leqslant (m+1) f(r).
$$

That is,
$$
\frac{m}{n} \leqslant \frac{f(s)}{f(r)} \leqslant \frac{m}{n} + \frac{1}{n}.
$$

Combining (1) and (2), we get
$$
\left|\frac{f(s)}{f(r)} - \frac{\ln s}{\ln r}\right| \leqslant \frac{1}{n}.
$$

Since $s, r, n \in \mathbf{N}^{+}$, for any given positive integers $s \geqslant r \geqslant 2$, by the arbitrariness in (3), we have

That is,
$$
\begin{array}{c}
\frac{f(s)}{f(r)} = \frac{\ln s}{\ln r}, \\
\frac{f(s)}{\ln s} = \frac{f(r)}{\ln r}. \\
f(s) = \frac{f(2)}{\ln 2} \ln s.
\end{array}
$$

In particular, taking $r = 2$, we have
By the arbitrariness of $s$, we know that for any positive integer $n \geqslant 2$,
$$
f(n) = \beta \ln n,
$$

where $\beta = \frac{f(2)}{\ln 2}$.
By the condition $f(1) = 0$, and $f(2) \geqslant f(1) \geqslant 0$, we know $\beta \geqslant 0$.
Note (1) When discussing solutions in the set of real numbers, the equation is usually solved through exponential transformation. However, this method does not work in the set of positive integers, because $e^{n}$ is not a positive integer.
(2) Let $f(n) = \ln g(n)$, by the conclusion we get:
Suppose $g$ is defined on the set of natural numbers $\mathbf{N}$. For any $m, n \in \mathbf{N}^{+}$, we have
$$
g(m n) = g(m) g(n),
$$

and $g(n+1) \geqslant g(n)$. Then $g(n) = n^{\beta}$, where $\beta = \frac{\ln g(2)}{\ln 2} = \log_{2} g(2)$.