Rectangle $ ABCD$ has area 2006.  An ellipse with area $ 2006\pi$ passes through $ A$ and $ C$ and has foci at $ B$ and $ D$.  What is the perimeter of the rectangle?  (The area of an ellipse is $ \pi ab$, where $ 2a$ and $ 2b$ are the lengths of its axes.)

$ \textbf{(A) } \frac {16\sqrt {2006}}{\pi} \qquad \textbf{(B) } \frac {1003}4 \qquad \textbf{(C) } 8\sqrt {1003} \qquad \textbf{(D) } 6\sqrt {2006} \qquad \textbf{(E) } \frac {32\sqrt {1003}}\pi$
1. Let \( AB = CD = x \) and \( BC = DA = y \). The area of the rectangle is given by:
   \[
   xy = 2006 \quad \text{(1)}
   \]

2. The area of the ellipse is given by:
   \[
   \pi ab = 2006\pi \implies ab = 2006 \quad \text{(2)}
   \]

3. Since \( B \) and \( D \) are the foci of the ellipse, the distance between them is \( 2c \). In a rectangle, the distance between opposite corners \( A \) and \( C \) is the diagonal, which is:
   \[
   AC = \sqrt{x^2 + y^2}
   \]
   Since \( A \) and \( C \) lie on the ellipse, the sum of the distances from any point on the ellipse to the foci is \( 2a \). Therefore:
   \[
   x + y = 2a \quad \text{(4)}
   \]

4. Using equation (4), we can express \( a \) as:
   \[
   a = \frac{x + y}{2} \quad \text{(5)}
   \]

5. From equation (2), we can express \( b \) as:
   \[
   b = \frac{2006}{a} = \frac{2006}{\frac{x + y}{2}} = \frac{4012}{x + y} \quad \text{(6)}
   \]

6. The distance between the foci \( B \) and \( D \) is \( 2c \), and since \( B \) and \( D \) are the foci of the ellipse, we have:
   \[
   c = \sqrt{\frac{x^2 + y^2}{4}} \quad \text{(7)}
   \]

7. In an ellipse, the relationship between \( a \), \( b \), and \( c \) is given by:
   \[
   a^2 = b^2 + c^2 \quad \text{(8)}
   \]

8. Substituting equations (5), (6), and (7) into equation (8), we get:
   \[
   \left( \frac{x + y}{2} \right)^2 = \left( \frac{4012}{x + y} \right)^2 + \left( \frac{\sqrt{x^2 + y^2}}{2} \right)^2
   \]

9. Simplifying the equation:
   \[
   \frac{(x + y)^2}{4} = \frac{4012^2}{(x + y)^2} + \frac{x^2 + y^2}{4}
   \]

10. Multiplying through by 4 to clear the denominators:
    \[
    (x + y)^2 = \frac{4 \cdot 4012^2}{(x + y)^2} + x^2 + y^2
    \]

11. Using the fact that \( x^2 + y^2 = 2xy \) from the Pythagorean theorem in the rectangle:
    \[
    (x + y)^2 = \frac{4 \cdot 4012^2}{(x + y)^2} + 2xy
    \]

12. Substituting \( xy = 2006 \) from equation (1):
    \[
    (x + y)^2 = \frac{4 \cdot 4012^2}{(x + y)^2} + 2 \cdot 2006
    \]

13. Simplifying further:
    \[
    (x + y)^2 = \frac{4 \cdot 4012^2}{(x + y)^2} + 4012
    \]

14. Let \( z = x + y \). Then:
    \[
    z^2 = \frac{4 \cdot 4012^2}{z^2} + 4012
    \]

15. Multiplying through by \( z^2 \):
    \[
    z^4 = 4 \cdot 4012^2 + 4012z^2
    \]

16. Rearranging the equation:
    \[
    z^4 - 4012z^2 - 4 \cdot 4012^2 = 0
    \]

17. Let \( u = z^2 \). Then:
    \[
    u^2 - 4012u - 4 \cdot 4012^2 = 0
    \]

18. Solving this quadratic equation using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
    \[
    u = \frac{4012 \pm \sqrt{4012^2 + 4 \cdot 4 \cdot 4012^2}}{2}
    \]
    \[
    u = \frac{4012 \pm \sqrt{4012^2 + 16 \cdot 4012^2}}{2}
    \]
    \[
    u = \frac{4012 \pm \sqrt{17 \cdot 4012^2}}{2}
    \]
    \[
    u = \frac{4012 \pm 4012\sqrt{17}}{2}
    \]
    \[
    u = 2006(1 \pm \sqrt{17})
    \]

19. Since \( u = z^2 \) and \( z = x + y \), we take the positive root:
    \[
    z = \sqrt{2006(1 + \sqrt{17})}
    \]

20. The perimeter of the rectangle is:
    \[
    P = 2(x + y) = 2z = 2\sqrt{2006(1 + \sqrt{17})}
    \]

The final answer is \( \boxed{8\sqrt{1003}} \).