jmc

From the given equations, $$\begin{array}{rl}(x-3)& =\pm (y-9),\\ (x-9)& =\pm 2(y-3).\end{array}$$Thus, we divide into cases.

Case 1: $x-3=y-9$ and $x-9=2(y-3).$

Solving this system, we find $(x,y)=(-15,-9).$

Case 2: $x-3=y-9$ and $x-9=-2(y-3).$

Solving this system, we find $(x,y)=(1,7).$

Case 3: $x-3=-(y-9)$ and $x-9=2(y-3).$

Solving this system, we find $(x,y)=(9,3).$

Case 4: $x-3=-(y-9)$ and $x-9=-2(y-3).$

Solving this system, we find $(x,y)=(9,3).$

Hence, the solutions $(x,y)$ are $(-15,-9),$ $(1,7),$ and $(9,3).$ The final answer is $(-15)+(-9)+1+7+9+3=\boxed{-4}.$