$4 \cdot 248$ To make the equation
$$
\left[\frac{10^{n}}{x}\right]=1989
$$

have integer solutions, what is the smallest value of the positive integer $n$?
[Solution] From the equation, we get
$$
1989 \leqslant \frac{10^{n}}{x} \cdot 0. \text{ Thus, from (1), we have }
$$
$$
10^{n} \cdot \frac{1}{1990}<x \leqslant 10^{n} \cdot \frac{1}{1989} \text {, }
$$

which means
$$
10^{n} \cdot 0.0005025 \cdots<x \leqslant 10^{n} \cdot 0.0005027 \cdots
$$

When $n=1,2,3, \cdots, 6$, there is no integer $x$ that satisfies (2), so the original equation has no integer solutions when $n<7$.
$$
\begin{array}{l}
\text { When } n=7, \text { (2) becomes } \\
5025 \cdots<x \leqslant 5027 . \cdots, \\
x_{1}=5026 \text { or } x_{2}=5027 .
\end{array}
$$

Upon verification, they are indeed the integer solutions of the original equation when $n=7$.
Therefore, the minimum value of $n$ for which the original equation has integer solutions is 7.