Employing multinomial expansion gives $$(x+y+z)^{2006}=\sum_{a+b+c=2006}\binom{2006!}{a!\cdot b!\cdot c!}x^ay^bz^c$$

By the basic counting pattern, there are $\binom{2006+2}{2}$ non-negative integer solutions to  the equation $a+b+c=2006$. This means there are $\binom{2008}{2}$ terms in the expansion of $(x+y+z)^{2006}$. Similarly, there are the same number of terms in the expansion of $(x-y-z)^{2006}$.

However, some terms in these two expansions will be cancelled. Such cancellations will occur when $b$ is odd and $c$ is even, or $b$ is even and $c$ is odd in the term $x^ay^bz^c$. These two scenarios are symmetric. Therefore, it is sufficient to just count the case when $b$ is odd and $c$ is even.

• When $b=1$, $c$ can take any even number between $0$ and $2004$. There are $1003$ possibilities.
• When $b=3$, $c$ can take any even number between $0$ and $2002$. There are $1002$ possibilities.
• $\cdots$
• When $b=2005$, $c$ can only take $0$. There is $1$ possibility.

Thus, the total count in this case is $$1+2+\cdots + 1002 + 1003= \frac{1003\times 1004}{2}$$

It follows that the final result is $$\binom{2008}{2}-\frac{1003\times 1004}{2}\times 2 = \boxed{1008106}$$