The expression
\[ (x \plus{} y \plus{} z)^{2006} \plus{} (x \minus{} y \minus{} z)^{2006}
\]is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$ \textbf{(A) } 6018 \qquad \textbf{(B) } 671,676 \qquad \textbf{(C) } 1,007,514 \qquad \textbf{(D) } 1,008,016 \qquad \textbf{(E) } 2,015,028$
1. **Rewrite the expression**: 
   We start with the given expression:
   \[
   (x + y + z)^{2006} + (x - y - z)^{2006}
   \]
   Notice that \((x - y - z)^{2006}\) can be rewritten as \(((-x) + y + z)^{2006}\) because raising to an even power negates the effect of the negative sign.

2. **Symmetry and cancellation**:
   When expanding \((x + y + z)^{2006}\) and \(((-x) + y + z)^{2006}\), each term in the expansion of \((x + y + z)^{2006}\) will have a corresponding term in the expansion of \(((-x) + y + z)^{2006}\). However, terms where the power of \(x\) is odd will cancel out because they will have opposite signs in the two expansions.

3. **Counting the terms**:
   We need to count the number of terms in the expansion of \((x + y + z)^{2006}\) where the power of \(x\) is even. 

4. **Using the multinomial theorem**:
   The number of terms in the expansion of \((x + y + z)^{2006}\) is given by the number of non-negative integer solutions to the equation:
   \[
   a + b + c = 2006
   \]
   where \(a\), \(b\), and \(c\) are the powers of \(x\), \(y\), and \(z\) respectively. This is a classic "stars and bars" problem, and the number of solutions is given by:
   \[
   \binom{2006 + 2}{2} = \binom{2008}{2}
   \]

5. **Calculating \(\binom{2008}{2}\)**:
   \[
   \binom{2008}{2} = \frac{2008 \times 2007}{2} = 2008 \times 1003 = 2016024
   \]

6. **Considering only even powers of \(x\)**:
   Since we are only interested in terms where the power of \(x\) is even, we substitute \(a = 2k\) where \(k\) is a non-negative integer. The equation becomes:
   \[
   2k + b + c = 2006
   \]
   Let \(m = 2006 - 2k\). Then \(b + c = m\) where \(m\) is even and ranges from 0 to 2006. The number of solutions for each \(m\) is:
   \[
   \binom{m + 1}{1} = m + 1
   \]
   Summing over all even \(m\) from 0 to 2006, we get:
   \[
   \sum_{k=0}^{1003} (2006 - 2k + 1) = \sum_{k=0}^{1003} (2007 - 2k)
   \]

7. **Summing the series**:
   \[
   \sum_{k=0}^{1003} 2007 - \sum_{k=0}^{1003} 2k = 1004 \times 2007 - 2 \sum_{k=0}^{1003} k
   \]
   The sum of the first 1004 natural numbers is:
   \[
   \sum_{k=0}^{1003} k = \frac{1003 \times 1004}{2} = 502506
   \]
   Therefore:
   \[
   1004 \times 2007 - 2 \times 502506 = 2016028 - 1005012 = 1008016
   \]

The final answer is \(\boxed{1008016}\).