$14 \cdot 25$ Find the smallest natural number $n$ such that the equation $\left[\frac{10^{n}}{x}\right]=1989$ has an integer solution $x$.
(23rd All-Soviet Union Mathematical Olympiad, 1989)
[Solution] Since
$$
\frac{10^{n}}{x}-1<\left[\frac{10^{n}}{x}\right] \leqslant \frac{10^{n}}{x},
$$

then
$$
\frac{10^{n}}{x}-1<1989 \leqslant \frac{10^{n}}{x} \text {, }
$$

which means
$$
\frac{10^{n}}{1990}<x \leqslant \frac{10^{n}}{1989} \text {. }
$$

Thus, we have
$$
10^{n} \cdot 0.000502512<x \leqslant 10^{n} \cdot 0.000502765 .
$$

From this inequality, we can see that only when $n=7$, we have
$$
5025.12<x \leqslant 5027.65 .
$$

At this point, the integer solution is $x=5026$.
Therefore, the smallest $n=7$.