4. (17th Nordic Mathematical Contest) Find all integer triples $(x, y, z)$ such that
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003 \text {. }
$$
4. $(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]=4006$.
Since $4006=2 \times 2003$, and $(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \equiv 0(\bmod 2)$,

Therefore, $\left\{\begin{array}{l}x+y+z=1, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=4006,\end{array}\right.$
or $\left\{\begin{array}{l}x+y+z=2003, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2 .\end{array}\right.$
For (1), we have
$$
(x-y)^{2}+(x+2 y-1)^{2}+(2 x+y-1)^{2}=4006 \text {, }
$$

which is $6 x^{2}+6 y^{2}+6 x y-6 x-6 y+2=4006$.
But $4006 \equiv 4(\bmod 6)$. Contradiction.
For (2), since $|x-y|,|y-z|,|z-x|$ have two 1s and one 0, without loss of generality, assume $x \geqslant y \geqslant z$.
When $x-1=y=z$, $3 y+1=2003$, no solution.
When $x=y=z+1$, $3 x-1=2003, x=668$.
Therefore, the integer solutions for the triplet are
$$
(668,668,667),(668,667,668),(667,668,668) .
$$