Find all four-digit natural numbers $\overline{xyzw}$ with the property that their sum plus the sum of their digits equals $2003$.
To find all four-digit natural numbers $\overline{xyzw}$ such that the sum of the number and the sum of its digits equals $2003$, we can follow these steps:

1. Let $\overline{xyzw}$ be a four-digit number, where $x, y, z, w$ are its digits. The number can be expressed as:
   \[
   1000x + 100y + 10z + w
   \]
   The sum of its digits is:
   \[
   x + y + z + w
   \]
   According to the problem, we have:
   \[
   1000x + 100y + 10z + w + x + y + z + w = 2003
   \]
   Simplifying this, we get:
   \[
   1001x + 101y + 11z + 2w = 2003
   \]

2. Since $\overline{xyzw}$ is a four-digit number, $x$ can only be $1$ or $2$ (as $x$ must be a non-zero digit and the number must be four digits).

3. **Case 1: $x = 2$**
   \[
   1001(2) + 101y + 11z + 2w = 2003
   \]
   Simplifying, we get:
   \[
   2002 + 101y + 11z + 2w = 2003
   \]
   \[
   101y + 11z + 2w = 1
   \]
   Since $101y + 11z + 2w = 1$ and $y, z, w$ are digits (0-9), there are no solutions in this case because the left-hand side is too large to equal 1.

4. **Case 2: $x = 1$**
   \[
   1001(1) + 101y + 11z + 2w = 2003
   \]
   Simplifying, we get:
   \[
   1001 + 101y + 11z + 2w = 2003
   \]
   \[
   101y + 11z + 2w = 1002
   \]

5. We need to find non-negative integer solutions for $y, z, w$ such that:
   \[
   101y + 11z + 2w = 1002
   \]
   Since $101y \leq 1002$, we have:
   \[
   y \leq \left\lfloor \frac{1002}{101} \right\rfloor = 9
   \]
   Therefore, $y$ can be $0, 1, 2, \ldots, 9$. However, we need to check which value of $y$ satisfies the equation.

6. Trying $y = 9$:
   \[
   101(9) + 11z + 2w = 1002
   \]
   \[
   909 + 11z + 2w = 1002
   \]
   \[
   11z + 2w = 93
   \]

7. We need to find non-negative integer solutions for $z$ and $w$ such that:
   \[
   11z + 2w = 93
   \]
   Trying $z = 7$:
   \[
   11(7) + 2w = 93
   \]
   \[
   77 + 2w = 93
   \]
   \[
   2w = 16
   \]
   \[
   w = 8
   \]

8. Therefore, the digits $x, y, z, w$ are $1, 9, 7, 8$ respectively, and the number is:
   \[
   \overline{xyzw} = 1978
   \]

The final answer is $\boxed{1978}$