Find those positive integers from which, when the sum of the digits of the number is subtracted, we get 2007.
Solution. Let the number be: $\overline{a_{1} a_{2} \ldots a_{n}}$. According to the problem statement:

$$
\begin{aligned}
\overline{a_{1} a_{2} \ldots a_{n}}-\left(a_{1}+a_{2}+\ldots+a_{n}\right) & =2007, \quad \text { that is } \\
a_{1}\left(10^{n-1}-1\right)+a_{2}\left(10^{n-2}-1\right)+\ldots+a_{n-1}(10-1) & =2007
\end{aligned}
$$

Since each term in the sum is non-negative, no term can be greater than 2007. Therefore, $a_{1}\left(10^{n-1}-1\right) \leq 2007$ must also hold, which implies $10^{n} \leq \frac{20070}{a_{1}}+10$. Since $a_{1} \geq 1$,

$$
\frac{20070}{a_{1}}+10 \leq 20080
$$

thus $10^{n} \leq 20080$ must also hold. Therefore, $n \leq 4$. However, $n>3$, because if we subtract the sum of the digits of a three-digit number, the result will be at most a three-digit number, always less than 2007.

Thus, $n=4$, meaning $\overline{a_{1} a_{2} a_{3} a_{4}}-\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=2007$,

$$
\begin{aligned}
& 999 a_{1}+99 a_{2}+9 a_{3}=2007 \\
& 111 a_{1}+11 a_{2}+a_{3}=223
\end{aligned}
$$

It is clear that $a_{1} \leq 2$, because if $a_{1}>2$, then the sum would be at least 333, which is greater than 223.

Case I: $a_{1}=2$. Then $222+11 a_{2}+a_{3}=223,11 a_{2}+a_{3}=1$. Here, $a_{2}<1$, because if $a_{2} \geq 1$, the sum would be greater than 1.

Thus, $a_{2}=0$, and $a_{3}=1$, while $a_{4} \in\{0,1,2,3, \ldots, 9\}$ can be arbitrary.

Case II: $a_{1}=1$. Then $111+11 a_{2}+a_{3}=223$, that is, $11 a_{2}+a_{3}=112$.

Since $a_{2} \leq 9$ and $a_{3} \leq 9$, $11 a_{2}+a_{3} \leq 108$. Therefore, there is no solution in this case.

The sought numbers are: 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019.