do we sum all the a's and b's and then say it cannot be equal to the a+b system's sum which is equivalent to the sum of the a's (all in mod m), because then m = m/2 (mod m) which is impossible?
Yes! That's the idea. Since each of $\sum a_k$ and $\sum b_k$ are $\dfrac{m}{2} \pmod{m}$, we have $\sum a_k + b_k = 0 \pmod{m}$, but we needed $\sum a_k + b_k = \dfrac{m}{2} \pmod{m}$.
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