From my subjective paper.....?

Knowing that $32^2 = 1024$ , $x, y \in 1, 2, ..., 31.$ Let $y = \sqrt{1009 - x^2}$ and working backwards from $x = 31:$

$y = \sqrt{1009 - 31^2} = \sqrt{48} = 4\sqrt{3};$

$y = \sqrt{1009 - 30^2} = \sqrt{109};$

$y = \sqrt{1009 - 29^2} = \sqrt{168} = 2\sqrt{42};$

$y = \sqrt{1009 - 28^2} = \sqrt{225} = 15$

Hence, the solution is $(x,y) = (28,15) \Rightarrow x + y = \boxed{43}.$