Easy as A B C

If $\frac{a^2 + b^2}{a^2 - b^2} = \frac{\sin(A + B)}{\sin(A - B)}$ , then by the sine addition formula $\frac{a^2 + b^2}{a^2 - b^2} = \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}$ , and after cross-multiplying this simplifies to $a^2 \cos A \sin B = b^2 \sin A \cos B$ , and after using the law of sines this simplifies to $\sin A \cos A = \sin B \cos B$ , which means $2 \sin A \cos A = 2 \sin B \cos B$ or $\sin 2A = \sin 2B$ . Since $a \neq b$ , $2A = \pi - 2B$ or $A + B = \frac{\pi}{2}$ , which means $C = \pi - (A + B) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$ . Therefore, $\triangle ABC$ is a right triangle.

This means $y = \frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C} = \frac{\sin A + \sin (\frac{\pi}{2} - A) + \sin \frac{\pi}{2}}{\cos A + \cos (\frac{\pi}{2} - A) + \cos \frac{\pi}{2}} = \frac{\sin A + \cos A + 1}{\cos A + \sin A + 0} = \frac{\sin A + \cos A + 1}{\sin A + \cos A}$ . The derivative of this is $y' = \frac{\sin A - \cos A}{(\sin A + \cos A)^2}$ , which when set equal to $0$ solves to $A = B = \frac{\pi}{2}$ for $y = \frac{\sin \frac{\pi}{2} + \cos \frac{\pi}{2} + 1}{\sin \frac{\pi}{2} + \cos \frac{\pi}{2}} = 1 + \frac{\sqrt{2}}{2}$ . The other extremes $A = 0$ and $A = \frac{\pi}{2}$ both solve to $y = \frac{\sin 0 + \cos 0 + 1}{\sin 0 + \cos 0} = 2$ .

Therefore the range of $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ is $(1 + \frac{\sqrt{2}}{2}, 2)$ , so that $p = 1 + \frac{\sqrt{2}}{2}$ , $q = 2$ , $p + q = 3 + \frac{\sqrt{2}}{2}$ and $\lfloor 10000(p + q) \rfloor = \boxed{37071}$ .