1.15 $\star \star$ For what minimum value of the natural number $n$ does the equation
$$
\left[\frac{10^{n}}{x}\right]=1989
$$

have an integer solution?
The smallest $n=7$. Clearly, $n \geqslant 4$, and $1989 \leqslant \frac{10^{n}}{x}<1990$, which means $\frac{10^{n}}{1990}<x \leqslant \frac{10^{n}}{1989}$. It can be verified that for $n=4,5,6$, the interval $\left(\frac{10^{n}}{1990}, \frac{10^{n}}{1989}\right]$ does not contain any integers. When $n=7$, $x=5026$ and 5027 are solutions to the original equation.