Interesting. Call the angle $\angle ECI = \alpha$ and $\angle FCI = \beta$. We want to find $\alpha + \beta$. Now $EF$ being the perp bisector of $CI$ implies $EC = EI$ and $FC = FI$, so then $\angle EIC = \alpha$ and $\angle FIC = \beta$. Now by the sum of exterior angles and isosceles triangles, we get $\angle AIE = \angle AEI = 2 \alpha$ and $\angle BIF = \angle BFI = 2 \beta$, which then implies $\angle EAI = 180 ^ \circ - 4 \alpha$ and $\angle FBI = 180 ^ \circ - 4 \beta$ But since $\angle EAI$ is exactly half of $\angle A$ and $\angle FBI$ is exactly half of $\angle B$, then we must have $$2(180 ^ \circ - 4 \alpha) + 2(180 ^ \circ - 4 \beta) + (\alpha + \beta) = 180 ^ \circ \Longrightarrow 720 ^ \circ - 7(\alpha + \beta) = 180 ^ \circ$$ Now solving for $\alpha + \beta$, we get $\frac{540}{7} ^ \circ$, so the answer is $\boxed {547}$, which happens to be the same answer as 2020 AIME I #1

@above the two problems aren't same, therefore it might be a coincidence.

in the above ZETA_in_olympia's post, the angle $\angle ECI = \alpha$ , $\angle FCI = \beta$ as EF is the perpendicular bisector, then by simple angle chasing $\alpha = \beta $