Find such a three-digit number that, when doubled, we get the number expressing the number of digits required to write all consecutive integers from one to this sought three-digit number (inclusive).

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Let $x$ be the desired number. There are 9 single-digit numbers, 90 two-digit numbers, and the number of three-digit numbers up to $x$ inclusive is $x-99$. The number of digits required to write all consecutive numbers from 1 to $x$ is $9 + 2 \cdot 90 + 3 \cdot (x - 99) = 3x - 108$. We obtain the equation $2x = 3x - 108$, from which $x = 108$.

## Answer

108.