Find all triples of integers ${(x, y, z)}$ satisfying ${x^3 + y^3 + z^3 - 3xyz = 2003}$
1. Given the equation \( x^3 + y^3 + z^3 - 3xyz = 2003 \), we can use the identity for the sum of cubes:
   \[
   x^3 + y^3 + z^3 - 3xyz = (x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)
   \]
   This transforms our equation into:
   \[
   (x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2) = 2003
   \]

2. Since \( x, y, z \) are integers, both \( x+y+z \) and \( (x-y)^2 + (y-z)^2 + (z-x)^2 \) must be integers. We need to find pairs of integers \((a, b)\) such that:
   \[
   a \cdot b = 2003
   \]
   and \( a = x+y+z \) and \( b = (x-y)^2 + (y-z)^2 + (z-x)^2 \).

3. The prime factorization of 2003 is:
   \[
   2003 = 2003 \times 1
   \]
   Therefore, the possible pairs \((a, b)\) are \((1, 2003)\) and \((2003, 1)\).

4. Consider the case \( x+y+z = 1 \) and \((x-y)^2 + (y-z)^2 + (z-x)^2 = 2003 \):
   \[
   (x-y)^2 + (y-z)^2 + (z-x)^2 = 2003
   \]
   Since 2003 is not a perfect square, there are no integer solutions for this case.

5. Consider the case \( x+y+z = 2003 \) and \((x-y)^2 + (y-z)^2 + (z-x)^2 = 1 \):
   \[
   (x-y)^2 + (y-z)^2 + (z-x)^2 = 1
   \]
   The only way for the sum of three squares to be 1 is if one of the squares is 1 and the other two are 0. This implies:
   \[
   (x-y)^2 = 1, \quad (y-z)^2 = 0, \quad (z-x)^2 = 0
   \]
   This means \( y = z \) and \( x = z \pm 1 \).

6. Without loss of generality, assume \( x \leq y \leq z \). Then:
   \[
   y = z \quad \text{and} \quad x = z - 1
   \]
   Substituting \( x = z - 1 \) and \( y = z \) into \( x + y + z = 2003 \):
   \[
   (z-1) + z + z = 2003
   \]
   \[
   3z - 1 = 2003
   \]
   \[
   3z = 2004
   \]
   \[
   z = 668
   \]
   Therefore:
   \[
   y = 668 \quad \text{and} \quad x = 667
   \]

7. The solutions are:
   \[
   (x, y, z) = (667, 668, 668), (668, 667, 668), (668, 668, 667)
   \]

The final answer is \( \boxed{ (667, 668, 668), (668, 667, 668), (668, 668, 667) } \)