Example 7 A four-digit number has the following property: dividing this four-digit number by its last two digits yields a perfect square (if the tens digit is zero, then divide by the units digit), and this perfect square is exactly the square of the first two digits plus 1. For example, $4802 \div 2=2401=49^{2}=$ $(48+1)^{2}$. The smallest four-digit number with the above property is $\qquad$
(1994, Sichuan Province Junior High School Mathematics League Final)
Solution: Let the first two digits of the four-digit number be $c_{1}$, and the last two digits be $c_{2}$. Then $10 \leqslant c_{1} \leqslant 99,1 \leqslant c_{2} \leqslant 99$. The four-digit number can be represented as $100 c_{1}+c_{2}$. According to the problem, we have
$$
\begin{array}{l}
100 c_{1}+c_{2}=\left(c_{1}+1\right)^{2} c_{2}=c_{1}^{2} c_{2}+2 c_{1} c_{2}+c_{2} . \\
\therefore 100 c_{1}=c_{1} c_{2}\left(c_{1}+2\right) . \\
\therefore c_{2}=\frac{100}{c_{1}+2} .
\end{array}
$$

Therefore, $\left(c_{1}+2\right) \mid 100$.
Noting that $10 \leqslant c_{1} \leqslant 99$,
$\therefore c_{1}$ can only be $18,23,48,98$.
Then $c_{2}$ is correspondingly $5,4,2,1$.
Thus, the four-digit numbers sought are $1805,2304,4802$, 9801. The smallest four-digit number among them is 1805.