The first 20 numbers of the natural number sequence were written on the board. When one of the numbers was erased, it turned out that among the remaining numbers, one is the arithmetic mean of all the others. Find all the numbers that could have been erased.

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The first method. The sum of the first twenty numbers of the natural series is equal to $(1+20) \cdot 10=210$. Therefore, if one of them is erased, the sum $S$ of the remaining numbers satisfies the inequality $190 \leq S \leq 209$. The arithmetic mean of the remaining numbers is $S / 19$. By the condition, this is a natural number, so $S$ is divisible by 19. There are exactly two such numbers in the interval $[190,209]$: 190 and 209. If $S=190$, then the number 20 was erased, and if $S=209$, then the number one was erased.

The second method. Notice that the largest or the smallest of the written numbers could have been erased. Indeed, in the first case, the arithmetic mean of the remaining numbers is ( $210-20$ ) : $19=10$, and in the second case, it is $(210-1): 19=11$. By erasing any other number, the arithmetic mean of the remaining numbers would be greater than 10 but less than 11, i.e., not an integer. Thus, no other number could have been erased.

## Answer

20 or 1.