23. (2005 Chinese National Team Training Selection Contest Problem) Let $\alpha$ be a given positive real number. Find all functions $f: \mathbf{N}^{+} \rightarrow \mathbf{R}$, such that for any positive integers $k$, $m$ satisfying $\alpha m \leqslant k<(\alpha+1) m$, we have $f(k+m)=f(k)+f(m)$.
23. $f(n)=b m, b$ is any given real number.

First, prove that when $n \geqslant 2 \alpha+3$, we have $f(n+1)-f(n)=f(n)-f(n-1)$.
It suffices to prove that there exists a positive integer $u$ such that
$$
\begin{array}{l}
f(n+1)-f(n)=f(u+1)-f(u), \\
f(n)-f(n-1)=f(u+1)-f(u) .
\end{array}
$$

Equation (1) is equivalent to
$$
f(n+1)+f(u)=f(n)+f(u+1) .
$$

By the given conditions, we only need
$$
\begin{array}{l}
\alpha(n+1) \leqslant u1$ when, take a positive integer $k$ such that
$$
a n_{1} \leqslant kn_{1}, k+n_{1}>n_{1}, f\left(k+n_{1}\right)=f(k)+f\left(n_{1}\right)$.
Thus,
$$
\begin{aligned}
f\left(n_{1}\right) & =f\left(k+n_{1}\right)-f(k) \\
& =\left(k+n_{1}\right) b-k b=n_{1} b,
\end{aligned}
$$

which is a contradiction.
When $\alpha \leqslant 1$, we have $\alpha n_{1} \leqslant n_{1}<(\alpha+1) n_{1}$. Therefore,
$$
\text { b. } 2 n_{1}=f\left(2 n_{1}\right)=f\left(n_{1}\right)+f\left(n_{1}\right) \text {. }
$$

Thus, $f\left(n_{1}\right)=n_{1} b$, which is a contradiction.
Furthermore, when $f(n)=b n, b$ is any given positive real number, it clearly satisfies the condition.
In summary, the required function is $f(n)=b m, b$ is any given real number.