jmc

Let $2a$ and $2b$ be the lengths of the major and minor axes of the ellipse, respectively, and let the dimensions of the rectangle be $x$ and $y.$ Then $x+y$ is the sum of the distances from the foci to point $A$ on the ellipse, which is $2a,$ so $x+y=2a.$ Also, the length of a diagonal of the rectangle is $\sqrt{x^2+y^2},$ which is also equal to the distance between the foci of the ellipse, which is $2\sqrt{a^2-b^2}.$ Thus, $x^2+y^2=4(a^2-b^2).$ Then the area of the rectangle is $$2006=xy=r\frac{1}{2}\left[(x+y{)}^2-(x^2+y^2)\right]=r\frac{1}{2}\left[(2a{)}^2-(4a^2-4b^2)\right]=2b^2,$$so $b=\sqrt{1003}.$ Thus, the area of the ellipse is $$2006\pi =\pi ab=\pi a\sqrt{1003}.$$Thus, $a=2\sqrt{1003},$ and the perimeter of the rectangle is $2(x+y)=4a=\boxed{8\sqrt{1003}}.$