Let's find the smallest positive integer whose square begins with 1989.
If $k \geq 0$ is an integer, then the smallest and the largest $(k+4)$-digit integer starting with 1989 is $1989 \cdot 10^{k}$ and $1990 \cdot 10^{k}-1$, respectively. We need to find the smallest positive integer $A$ for which there exists a $k \geq 0$ such that

$$
1989 \cdot 10^{k} \leq A^{2} < 1990 \cdot 10^{k}
$$

or, equivalently, the non-negative integer $k$ for which the

$$
\left[\sqrt{1989} \cdot (\sqrt{10})^{k} ; \quad \sqrt{1990} \cdot (\sqrt{10})^{k}\right)
$$

left-closed, right-open interval contains an integer $(A)$.

From the following inequalities, it follows that this first occurs for $k=2$.

$$
\begin{aligned}
& k=0: \quad 44 < \sqrt{1989} < \sqrt{1990} < 45 \\
& k=1: \quad 141 < \sqrt{1989} \cdot \sqrt{10} < \sqrt{1990} \sqrt{10} < 142 \\
& k=2: \quad 445 < 10 \sqrt{1989} < 446 < 10 \sqrt{1990}
\end{aligned}
$$

Thus, the number we are looking for is 446; its square is 198916.

Remark. Two students, Stöhr Lóránt and Wiener Gábor, also examined the solution of the problem in other number bases and found that in the base-28 number system, 1989 itself is a perfect square, and thus the smallest solution is

$$
63_{28}^{2}=1989_{28}
$$

resulting in $63_{28}$.

Based on the work of Katz Sándor (Bonyhád, III. sz. Ált. Isk., 8. o. t.)