Rewriting the grid in base 8,
I claim that at this point, it is essentially proven. If I choose any set of numbers as described in the question, the numbers I select must contain exactly one every last digit(0-7), and exactly one of every second digit 0-7. Note that last digit of 0 would imply that it is no longer in the same residue since the numbers start from 1, so the number chosen that last digit of 0 bears an extra value of 8 rather than 0. This means the sum of the numbers chosen are, $$S=(70+60+50+40+30+20+10+1+2+3+4+5+6+7+10)_8$$ $$S=(8(7+6+5+4+3+2+1) + 1+2+3+4+5+6+7+8)_{10}$$ $$S=260_{10} \ \blacksquare$$
(Please confirm the validity of my proof, it's very different from the given answer to the question)
Each numbercan be represented as $8r+c$ where $c$ is the column number and $r$ is the row number.(starting from 0) One number must be selected from every row, this means that every distinct $r$ and distinct $c$ must be selected.
$$\frac{8 \cdot 9}{2} + \frac {8 \cdot(7\cdot 8)}{2}=36+224=260$$
So your proof is valid.