Alice chooses the first number on the board $\not\equiv1\pmod{17}$ and squares it continuously until it becomes $1\pmod{17}$ regardless of what Bob does and then moves on to the next number $\not\equiv1\pmod{17}$ and repeats this until she has done this to all numbers $\not\equiv1\pmod{17}$ on the board. proofNote that the Quadratic residues of 17 form the set $\pm1,\pm2\pm4\pm8$ so if Alice keeps squaring a number it will eventually be $1\pmod{17}$ moreover note that if Bob cubes a number which Alice squared in the previous move then since the residues of $n^6$ form the exact same set $\pm1,\pm2\pm4\pm8$ Bob can do nothing to stop Alice from making a number $1\pmod{17}$ (in the next move of Alice that number has to be $\pm1\pm4\pmod{17}$ so we can guarantee it) hence making the sum of all numbers divisible by 17.