Solution. If $x=0$, then $20023=2023 \cdot 9+1816$, which means 20023 is not divisible by 2023.

Let the number $x \in\{1,2, \ldots, 9\}$. Then we have

$$
\overline{20 x 23}=20000+\overline{x 00}+23=2023 \cdot 10-230+x \cdot 100+23=2023 \cdot 10+x \cdot 100-207
$$

Since $-107 \leq x \cdot 100-207<2023$ and $x \cdot 100-207 \neq 0$, it follows that for any digit $x \in\{1,2, \ldots, 9\}$, the number $\overline{20 \times 23}$ is not divisible by 2023.

Let the number $x=\overline{m n}=10 m+n, m \in\{1,2, \ldots, 9\}, n \in\{0,1,2, \ldots, 9\}$. Then we have

$$
\begin{gathered}
\overline{20 \times 23}=\overline{20 m n 23}=200000+\overline{m n 00}+23=202300-2300+\overline{m n} \cdot 100+2023-2000= \\
2023 \cdot 101+(\overline{m n}-43) \cdot 100
\end{gathered}
$$

We have $2023=7 \cdot 17^{2}$, and $100=2^{2} \cdot 5^{2}$. It follows that the natural numbers 2023 and 100 are coprime.

If $10 \leq \overline{m n}<43$, then $\overline{m n}-43$ cannot be divisible by 2023, since $0<43-\overline{m n}<2023$. Therefore, the number $\overline{20 \times 23}$ is not divisible by 2023.

If $\overline{m n}=43$, then $204323=2023 \cdot 101$. Thus, $x=43$ is the smallest natural number with the property stated in the problem.
8.2. On the board, a natural number $x$ is written. Before the number $x$, we write the digits 20, and after it, we write the digits 23 (for example, if $x=1957$, then the number 20195723 is obtained). Determine the smallest natural number $x$ with the property that the obtained number is divisible by 2023.