Problem 9.1. Find all natural numbers $n(n>1)$ that satisfy the following condition: from the set of numbers $\{1,2,3, \ldots, n\}$, a number can be removed such that the arithmetic mean of the numbers in the set changes by $\frac{1}{2020}$. For each such number $n$, also identify the number removed.
Solution. Let $n$ be such a number and $A=\{1,2,3, \ldots, n\}$. The sum of the numbers in set $A$, $S=\frac{n(n+1)}{2}$, and their mean, $M=\frac{n+1}{2}$. Let $m$ be the number removed from set $A$, $1 \leq m \leq n$. Let $A^{\prime}$ be the set of remaining elements, $A^{\prime}=A \backslash\{m\}=\{1,2, \ldots, m-1, m+1, \ldots, n\}$. The sum of the numbers in set $A^{\prime}$, $S^{\prime}=S-m=\frac{n(n+1)-2 m}{2}$, and their mean, $M^{\prime}=\frac{n(n+1)-2 m}{2(n-1)}$. As a result of removing the number $m$, the mean of the numbers can increase or decrease by $\frac{1}{2020}$. According to the condition,

$M-M^{\prime}=\frac{1}{2020}$ or $M^{\prime}-M=\frac{1}{2020}$. Both cases are examined.

Case 1. Let $M-M^{\prime}=\frac{1}{2020}$. We obtain $\frac{n+1}{2}-\frac{n(n+1)-2 m}{2(n-1)}=\frac{2 m-n-1}{2(n-1)}=\frac{1}{2020}$, from which $n=\frac{2020 m-1009}{1011}=\frac{(2022 m-1011)-2(m-1)}{1011}=2 m-1-2 \cdot \frac{m-1}{1011}$. Since $n$ is a natural number, and $(2,1011)=1$, it follows that $\frac{m-1}{1011}$ must be a natural number. Let $\frac{m-1}{1011}=k$, where $k \in \mathrm{N}$. Thus, $m=1011 k+1$. But then $n=2(1011 k+1)-1-2 \cdot k$, that is, $n=2020 k+1$. Since $m \leq n$, it follows that $k>0$, that is, $k \in \mathrm{N}^{*}$.

Thus, in Case 1, the numbers $n$ are all numbers of the form $n=2020 k+1$; the number removed from set $A$ is $m=1011 k+1$.

Case 2. Let $\quad M^{\prime}-M=\frac{1}{2020}$. As in Case 1, it is obtained that the numbers $n$ are all numbers of the form $n=2020 k+1$; the number removed from set $A$ is $m=1009 k+1$.

Definitely, the condition in the statement is satisfied by all natural numbers of the form $n=2020 k+1$, where $k=1,2,3, \ldots$. The respective numbers removed from $A$ are numbers of the form $m=1011 k+1$ and $m=1009 k+1$.