The one ($A$) who arrives at $1$ loses. So the one ($B$) who arrives at $2$ wins, because he subtracts one and his opponent arrives at $1$. Consequently, $A$ must be at $3$ on the series, because he only can subtract $1$, so $B$ arrives at $2$. If it were the other way $A$ would win, because he could follow the same way. Therefore $B$ must be on series at $4$, so he can bring $B$ to $3$, then $B$ subtract $1$ and brings $A$ to $2$, then $A$ subtract $1$ and $B$ loses. Thus $A$ must be on series at $5$, so he has to subtract $1$. And so on $....$
So $A$, the one who loses always must to be on series at odd numbers and $B$, the one who wins should be on series at divisible numbers.
Ann starts with a divisble number, so if she plays right, she should win.
Ann always puts Bob on an odd number $(2 * n + 1)$. So Bob only can subtract odd divisors $(2 * k + 1)$. Therefore Ann always arrives at a number divisible by $2$, because:
$(2 * n + 1) - (2 * k + 1) = 2 * n - 2 * k = 2 * (n - k)$
So it is impossible for Bob to reach $2$, if Ann plays good, he can't win.
1) The optimal strategy is to...
always leave the other with odd numbers.
2) If both are playing optimal...
Ann will win.
If $N$ is odd Bob will win, else Ann will win.
Bob, Charles, David and Edward will connect against Ann and will try to kick Ann out, because all of them now, that if Ann will play with the strategy from point 2) they will lose in the final. In the next game Charles, David and Edward will connect against Bob, because of the same reasons as before. And so on$... $
Eduard know that he can't win the whole game, in every final he will only has the second turn and can't win against the strategy from point 2). So David will win the game. If the start number $N$ isn't $60$ rather any number, you can't say who will win.