Example 39 (17th Nordic Mathematical Contest) Find all integer triples $(x, y, z)$ such that $x^{3}+y^{3}+z^{3}-3 x y z=2003$.
From the identity $a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ $=\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$, we get
$(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]=4006=2 \times 2003$, and $(x-y)^{2}+$ $(y-z)^{2}+(z-x)^{2}$ must be even, hence
(1) $\left\{\begin{array}{l}x+y+z=2003, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2,\end{array}\right.$
or (2) $\left\{\begin{array}{l}x+y+z=1, \\ (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=4006 \text {. }\end{array}\right.$

For (1), since $|x-y|,|y-z|,|z-x|$ have 2 values of 1 and 1 value of 0, without loss of generality, assume $x \leqslant y \leqslant z$, then $x-1=y=z$ or $x=y=z+1$. When $x-1=y=2$, $3 y+1=2003$ has no solution; when $x=$ $y=z+1$, $3 x-1=2003, x=668$. By symmetry, there are three solutions $(668,668,667),(668$, $667,668),(667,668,668)$.
For (2), we have $(x-y)^{2}+(x+2 y-1)^{2}+(2 x+y-1)^{2}=4006$,

which simplifies to $6 x^{2}+6 y^{2}+6 x y-6 x-6 y+2=4006$, leading to $2 \equiv 4(\bmod 6)$, a contradiction.
Therefore, the 3-tuples $(x, y, z)$ that satisfy the condition are $(668,668,667),(668,667,668),(667$, $668,668)$.