Three, (Full marks 50 points) If the hundreds digit of an $n$-digit natural number $N$ is 9, and the sum of its digits is $M$, where $n>3$, when the value of $\frac{N}{M}$ is the smallest, what is $N$?

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$$
\begin{array}{l}
\text { Three, (1) When } n=4 \text {, let } N=\overline{a 9 b c} \text {. Then } \\
\frac{N}{M}=\frac{1000 a+900+10 b+c}{a+9+b+c} \\
=1+\frac{999 a+891+9 b}{a+9+b+c} \\
\geqslant 1+\frac{999 a+891+9 b}{a+9+b+9} \text { (equality holds when } c=9 \text {) } \\
=1+\frac{9(a+b+18)+990 a+729}{a+b+18} \\
=10+\frac{990 a+729}{a+b+18} \\
\geqslant 10+\frac{990 a+729}{a+9+18} \text { (equality holds when } b=9 \text {) } \\
=10+\frac{990(a+27)-20 \text { aip1 }}{a+27} \\
=1000-\frac{26001}{a+27} \\
\geqslant 1000-\frac{26001}{1+27} \text { (equality holds when } a=1 \text {) } \\
=\frac{1999}{28} \text {. } \\
\end{array}
$$

Therefore, when $\frac{N}{M}$ reaches its minimum value, $N$ is 1999.
(2) When $n>4$, prove by mathematical induction.
$$
10^{n-1}-648 n>0 \text {. }
$$
$1^{\circ}$ When $n=5$, it is obviously true.
$2^{\circ}$ Assume that when $n=k$, the proposition is true. That is,
$$
10^{k-1}-648 k>0 \text {. }
$$

Then, when $n=k+1$, we have
$$
\begin{array}{l}
10^{(k+1)-1}-648(k+1) \\
=10 \times 10^{k-1}-648 k-648 \\
=10^{k-1}-648 k+9 \times 10^{k-1}-648 \\
>9 \times 10^{k-1}-648>9 \times 10^{4}-648>0 .
\end{array}
$$

Therefore, when $n=k+1$, the proposition is true.
Thus, when $n>4$, $10^{k-1}-648 n>0$ always holds.
Hence, $\frac{N}{M} \geqslant \frac{10^{n-1}}{9 n}>\frac{648 n}{9 n}=72>\frac{1999}{28}$.
Therefore, when $\frac{N}{M}$ reaches its minimum value $\frac{1999}{28}$, $N$ is 1999.
(Xie Wenxiao, Huanggang High School, Hubei Province, 436100)
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