$4 \cdot 248$ 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
$$1989 \leqslant \frac{10^{n}}{x} < 1990$$. Thus, from (1), we have
$$10^{n} \cdot \frac{1}{1990}<x \leqslant 10^{n} \cdot \frac{1}{1989}$$

That is,
$$10^{n} \cdot 0.0005025 \cdots<x \leqslant 10^{n} \cdot 0.0005027 \cdots$$

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

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.