2007 AMC 12A Problem 19

The altitude $h$ from $A$ in $\triangle ABC$ satisfies $2007=\tfrac12\cdot 223\cdot h,$ so $h=18.$ Thus $A$ lies on $y=18$ or $y=-18.$ Line $DE$ has equation $x-y-300=0.$ The condition on $\triangle ADE$ similarly places $A$ on one of two lines parallel to $DE.$ The four possible positions of $A$ are the vertices of a parallelogram whose center is the intersection of $y=0$ with line $DE,$ namely $(300,0).$ Hence the sum of the four $x$-coordinates is $4\cdot 300=1200.$ Thus, the correct answer is E.