Six. (20 points) A company plans
to donate $N$ million yuan (where $130<N<150$) to $n$ Hope Primary Schools. The allocation method is as follows: give the first school 4 million yuan and $\frac{1}{m}$ of the remaining amount; give the second school 8 million yuan and $\frac{1}{m}$ of the new remaining amount; give the third school 12 million yuan and $\frac{1}{m}$ of the further remaining amount; and so on, giving the $k$-th school $4k$ million yuan and $\frac{1}{m}$ of the remaining amount, until the $n$-th school receives the last portion, and all schools receive the same amount. Determine the total donation amount $N$, the number of schools $n$, and the donation amount each school receives. (Where $N$, $n$, $m$, and $k$ are all positive integers.)
Six, let each school receive $x$ ten thousand yuan.
From the problem, we have $x=\frac{N}{n}$.
Also, since the $n$-th school receives $x=4n$, with the remainder being zero,
$$
\begin{array}{l}
\therefore N=(4n) \times n=4n^2. \\
\text{Also, since } 130<N<150,
\end{array}
$$

i.e., $130<4n^2<150$, and $n$ is an integer,
$$
\therefore 4n^2=144,
$$

i.e., $N=144, n=6, x=\frac{144}{6}=24$.
Answer: The total donation amount $N$ is 144 (ten thousand yuan), the number of schools is 6, and each school receives a donation of 24 ten thousand yuan.
(Provided by Ren Chaoyan, Taiyuan Educational Institute)