Example 3 If the digits of a four-digit number are reversed to form a new four-digit number, the new number is exactly four times the original number. Find the original number.
(1988, Nanjing Mathematical Olympiad Selection Contest)
Solution: Let the required four-digit number be $\overline{a b c d}$, then the new four-digit number is $\overline{d c b a}$. According to the problem, we have
$\overline{a b c d}+\overline{d c b a}=5 \overline{a b c d}$.
Obviously, the last digit of $a+d$ can only be 0 or 5.
$\because 4 a \leqslant d \leqslant 9$, and $a, d$ cannot be 0,
$\therefore a=1$ or $a=2$.
When $a=1$, $d$ can only be 4 or 9, and must satisfy
$\overline{4 c b 1}=4 \times \overline{1 b c 4}$,
$\overline{9 c b 1}=4 \times \overline{1 b c 9}$.
Considering the last digit, it is clearly not possible.
When $a=2$, $d$ can only be 3 or 8.
Obviously, $d=3$ does not hold.
When $d=8$, it must satisfy $\overline{8 c b 2}=4 \times \overline{2 b c 8}$.
Considering the last digit that can make the above equation true.
Further find $b, c$.
$\because 4 b \leqslant c \leqslant 9$ (obviously $b, c$ cannot be 0),
$\therefore b=1$ or $b=2$, and the last digit of $4 c+3$ should be the same as $b$.
When $b=2$, $c=8$. But the last digit of $4 \times 8+3$ is not 2.
When $b=1$, $c$ can be $4,5,6,7,8$. It is easy to know that the last digit of $4 \times 7 +3$ is 1. Thus,
$$
b=1, c=7 \text{. }
$$

At this point, $8712=4 \times 2178$.
Therefore, the required four-digit number is 2178.