Let's determine the smallest natural number in which the sum of the digits is 1981.
First, let's examine the minimum number of digits a natural number must have for the sum of its digits to be 1981. Since the maximum value of a digit is 9, the sum of the digits is at most nine times the number of digits. This means that the number of digits cannot be less than one-ninth of the sum of the digits.

In our case, \(1981 \div 9 = 220.1\), and the number of digits must be an integer, so it must be at least 221. Therefore, it is sufficient to select the smallest number from the 221-digit numbers with the given property, as this will be smaller than any number with more than 221 digits.

It is sufficient to examine the numbers where the first digit is as small as possible. The first digit is the smallest when the sum of the remaining 220 digits is as large as possible. This sum is at most \(220 \times 9 = 1980\), so the smallest value of the first digit is 1, and this is only possible if all the other digits are 9. Therefore, there is exactly one 221-digit number where the sum of the digits is 1981 and the first digit is 1. Any other number with the given property is either at least 222 digits long or 221 digits long but has a first digit of at least 2.

Thus, the sought number is:

![](https://cdn.mathpix.com/cropped/2024_05_02_b739261f0b5f57466351g-1.jpg?height=92&width=195&top_left_y=594&top_left_x=954)

Remark. Using a similar line of reasoning, we can determine that for a given \(n\), the smallest natural number whose digits sum to \(n\) is:

\[
r \underbrace{999 \ldots 9}_{k \text{ nines }} \quad \text{where} \quad n = k \cdot 9 + r, \quad 0 \leq r < 9
\]