From a natural number, the sum of its digits was subtracted and the result was 2007. What could the original number be?

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It is not difficult to verify that the desired number must be a four-digit number. Let it be $\overline{a b c d}$ $=1000 a+100 b+10 c+d$. Then $1000 a+100 b+10 c+d-(a+b+c+d)=999 a+99 b+9 c=9(111 a+11 b+c)$. We obtain the equation: $9(111 a+11 b+c)=2007 \Leftrightarrow 111 a+11 b+c=223$. By trial, we can see that $a>1$ and $a<3$, that is, $a=2$.

Then $11 b+c=1$. Since $b$ and $c$ are digits, then $b=0$ and $c=1$. Note that $d$ can be any digit.

## Answer

any natural number from 2010 to 2019.