Solution:
If $x \in (-\infty, -2007)$, then
$$-(x-2007)-(x+2007)=k \quad \text{or} \quad x=-\frac{k}{2}.$$
If $x \in [-2007, 2007]$, then
$$-(x-2007)+(x+2007)=k \quad \text{or} \quad k=4014.$$
If $x \in (2007, +\infty)$, then
$$(x-2007)+(x+2007)=k \quad \text{or} \quad x=\frac{k}{2}.$$
Thus, the given equation has $(-\infty, -2007) \cup (2007, +\infty)$ as its solution set if and only if
$$\frac{k}{2} > 2007 \quad \text{or} \quad k > 4014.$$