1.40 Let $n$ be a five-digit number (the first digit is not zero), and $m$ be the four-digit number formed by removing the middle digit of $n$. Determine all $n$ such that $\frac{n}{m}$ is an integer.


[Solution] Let $n=\overline{x y z u v}=x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v$. Where $x, y, z, u, v$ are digits $0,1,2, \cdots, 8$ or $9, x \geqslant 1$.
$$
m=\overline{x y u v}=x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v \text {, }
$$

We first prove that $9 m<n<11 m$.
In fact, the inequality $9 m<n$, that is
$$
\begin{array}{l}
9\left(x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v\right) \\
\quad<x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v, \text { which is equivalent to } \\
80 u+80 v<10^{3} \cdot x+10^{2} \cdot y+10^{2} \cdot z,
\end{array}
$$

This is obviously correct.
The inequality $n<11 m$, that is
$$
\begin{array}{l}
x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v \\
<11\left(x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v\right)
\end{array}
$$

It is equivalent to
$$
10^{2} \cdot z<10^{3} \cdot x+10^{2} \cdot y+10^{2} u+10 v,
$$

This is obviously correct.
Thus, $9 m<n<11 m$ holds. Let $k=\frac{n}{m} \in \mathbb{N}$, then $n=k m$, so we get $9<k<11$. Since $k$ is an integer, so $k=10$, thus $n=10 m$, that is
$$
\begin{aligned}
& x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v \\
= & 10\left(x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v\right),
\end{aligned}
$$

or $z \cdot 10^{2}=90 u+9 v=9(10 u+v)$.
Since 9 can divide $z \cdot 10^{2}$, but 3 cannot divide $10^{2}$, so $z=9 t$ (where $t$ is an integer), thus
$$
t \cdot 10^{2}=10 u+v .
$$

From this we get $t=u=v=0$, hence we deduce
$$
u=\overline{x y 000}=N \cdot 10^{3}, 10 \leqslant N \leqslant 99 .
$$