Problem 815

[Solution] If the fractions $\frac{1}{n}$ and $\frac{1}{n+1}$ can be expressed as finite decimal numbers, then it must be that
$$\left\{\begin{array}{l} n=2^{a} \cdot 5^{b}, \\ n+1=2^{c} \cdot 5^{d} . \end{array}\right.$$

where $a, b, c, d$ are non-negative integers.
Thus, we have $\quad 2^{c} \cdot 5^{d}-2^{a} \cdot 5^{b}=1$.
When $a \leqslant c, \quad b \leqslant d$, we have
$$2^{a} \cdot 5^{b}\left(2^{c-a} \cdot 5^{d-b}-1\right)=1 .$$

That is
$$\left\{\begin{array}{l} 2^{a} \cdot 5^{b}=1, \\ 2^{c-a} \cdot 5^{d-b}-1=1 . \end{array}\right.$$

From this, we solve $a=0, b=0, c=1, d=0$.
Thus, $n=1, n+1=2$.
When $c \leqslant a, \quad b \leqslant d$, we have
$$2^{c} \cdot 5^{b}\left(5^{d-b}-2^{a-c}\right)=1 \text {. }$$

That is
$$\left\{\begin{array}{l} 2^{c} \cdot 5^{b}=1, \\ 5^{d-b}-2^{a-c}=1 . \end{array}\right.$$

From this, we solve $c=0, \quad b=0$, thus
$$5^{d}-2^{a}=1 .$$

Next, we find all non-negative integer solutions to (1).
Considering $\bmod 3$, we have
$$\begin{array}{l} 2^{2 k} \equiv 1 \quad(\bmod 3) \\ 2^{2 k+1} \equiv 2 \quad(\bmod 3) \end{array}$$

where $k$ is a non-negative integer.
Thus, from
$$1=5^{d}-2^{a} \equiv 2^{d}-2^{a}(\bmod 3),$$

we know that $d$ is odd, and $a$ is even, so
$$\begin{aligned} 2^{a} & =5^{d}-1 \\ & =(5-1)\left(5^{d-1}+5^{d-2}+\cdots+1\right) . \\ & 5^{d-1}+5^{d-2}+\cdots+1 \end{aligned}$$

Notice that
is the sum of $d$ odd numbers, and is a factor of $2^{a}$, so it can only be 1, thus $d=1$, and hence $a=2$.
That is, the non-negative integer solutions to (1) are only $a=2, d=1$.
At this time
$$n=2^{a} \cdot 5^{b}=4 \text {. }$$

Therefore, only
$$n=1 \text { and } n=4 \text {. }$$