Example 9 The decimal representation of a natural number $x$ does not
contain the digit 0, and it satisfies $x \cdot \bar{x}=1000+\Pi(x)$, where $\bar{x}$ is the natural number obtained by writing the digits of $x$ in reverse order, and $\Pi(x)$ is the product of the digits of $x$. Find all such natural numbers $x$.
(1991, Leningrad Mathematical Olympiad Third Round, Grade 8)
Solution: From the problem, we know that $x$ cannot be a one-digit number.
If $x$ is a two-digit number, let $x=10a+b$.
Then $\bar{x}=10b+a$
(where $a, b$ are natural numbers less than 10),
and it satisfies
$$
(10a+b)(10b+a)=1000+ab, \text{ (1) }
$$

which simplifies to $10ab+a^2+b^2=100$.
$$
\begin{array}{l}
\text { Also, since } a^2+b^2 \geqslant 2ab, \\
\therefore 12ab \leqslant 100 .
\end{array}
$$

Therefore, $ab \leqslant 8$.
Also, $a^2+b^2$ must be divisible by 10,
so $a, b$ can only be $(1,3)$, $(1,7)$, $(2,4)$. Substituting these into (1), only $(2,4)$ satisfies the equation.

If $x$ is a two-digit number, and $x$ does not contain 0 in its decimal representation, then
$$
\begin{aligned}
x \cdot \bar{x} & >100x>90x+1000>10x+1000 \\
& >1000+\Pi(x) .
\end{aligned}
$$

Therefore, $x$ cannot be a three-digit number or more. Thus, the natural numbers that satisfy the problem are 24 or 42.