Solution. The sum of all natural numbers from 1 to 2021 is

$$S=\frac{2021 \cdot 2022}{2}=2021 \cdot 1011$$

Let $x-1, x, x+1$ be the deleted numbers. Then the arithmetic mean of the remaining numbers is

$$N=\frac{2021 \cdot 1011-(x-1+x+x+1)}{2021-3}=\frac{2021 \cdot 1011-3 x}{2018}$$

We introduce the substitution $x=1011+y$, where $-1009 \leq y \leq 1009$ and get

$$N=\frac{2021 \cdot 1011-3(1011+y)}{2018}=\frac{2018 \cdot 1011-3 y}{2018}=1011-\frac{3 y}{2018} .$$

According to the condition of the problem, $N$ is a natural number, and from the constraint on $y$ we additionally have $-\frac{3}{2} \leq \frac{3 y}{2018} \leq \frac{3}{2}$. The integers in the interval $\left[-\frac{3}{2}, \frac{3}{2}\right]$ are $-1,0,1$, from which it follows that $y=0$ is the only possible value for $y$. Therefore, the deleted numbers are 1010, 1011, and 1012.