Six, 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 the condition
$$
\alpha m \leqslant k < (\alpha+1) m
$$

we have
$$
f(k+m)=f(k)+f(m).
$$
(Proposed by Yonggao Chen)
Six $f(n)=b n, b$ is any given real number.
First, prove: when $n \geqslant 2 \alpha+3$, we have
$$
f(n+1)-f(n)=f(n)-f(n-1) \text {. }
$$

It is sufficient 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
$$
f(n+1)+f(u)=f(n)+f(u+1) \text {. }
$$

From the given conditions, we only need
$$
\begin{array}{l}
\alpha(n+1) \leqslant u1 \text{ when } n \geqslant 2 \alpha+3.
\end{array}
$$

When $\alpha > 1$, take a positive integer $k$, such that
$\alpha n_{1} \leqslant kn_{1}, k+n_{1}>n_{1}, f\left(k+n_{1}\right)=f(k)+f\left(n_{1}\right)$.
Therefore, $f\left(n_{1}\right)=f\left(k+n_{1}\right)-f(k)$
$$
=\left(k+n_{1}\right) b-k b=n_{1} b,
$$

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

Thus, $f\left(n_{1}\right)=n_{1} b$, which is a contradiction.
Moreover, when $f(n)=b n, b$ is any given real number, it clearly satisfies the condition.

In conclusion, the required function is $f(n)=b n, b$ is any given real number.
(2000 China Mathematical Olympiad Committee provided)