Alice wins when $N$ is not a power of 2.
Let $i(N)$ be the maximal $i$ such that $N$ is divisble by $2^i$.
Alice's strategy: If there are $m$ stones left, Alice picks $2^{i(m)}$ stones.
Let me explain why this strategy is winning for Alice. Assume that Bob won by picking $b > 0$ remaining stones. Assume that before that Alice picked $2^j$ stones. By definition $j = i(2^j + b)$. Hence $2^j + b$ is divisible by $2^j$. Therefore $b$ is also divisible by $2^j$. This means that $b\ge 2^j$. On the other hand by the rules of the game $b\le 2^j$. Therefore $b = 2^j$.
But then $i(2^j + b) = i(2^j + 2^j) = j + 1$, contradiction.
This strategy is correct because (a) since $N$ is not a power of 2, Alice picks less than $N$ stones in the first turn; (b) Alice never picks more stones than Bob in the previous turn. Indeed, assume that Bob picked $b$ stones, before Alice picked $2^j$ stones, and we are left with $m > 0$ stones.
Let us verify that $2^{i(m)} \le b$. We will do it by showing that $b$ is divisble by $2^{i(m)}$.
Indeed, by definition of Alice's strategy $j = i(2^j + b + m)$ and by the rules of the game $b\le 2^j$. Let us show that $i(m) \le j$. Indeed, if $i(m) > j$, then $b$ is divisble by $2^j$, since both $2^j + b + m$ and $m$ are divisble by $2^j$. But since $b\le 2^j$, this means that $b = 2^j$. This contradicts the fact that $j = i(2^j + b + m)$. Indeed, since $i(m) > j$, we have that $i(2^j + b + m) = i(2^{j + 1} + m) \ge j + 1$.
Thus we have proved that $i(m) \le j$. This means $2^j + b + m$ is divisble by $2^{i(m)}$, as well as $m$. Hence $b$ is also divisble by $2^{i(m)}$ as required.
Bob wins when $N$ is a power of 2. Assume that $N = 2^i$ and Alice picks $a$ stones. Then Bob can use Alice's strategy described above. We only have to check that Bob then picks at most $a$ stones. Indeed, assume that $j$ is such that $2^i - a$ is divisible by $2^j$. Then $a$ is also divisible by $2^j$ hence $2^j \le a$.
Best Answer
Your game with piles of $S_1,\ldots,S_N$ stones is identical to the Nim game with piles of $S_1, \Bigl\lfloor \frac{S_2}2\Bigr\rfloor,\Bigl\lfloor \frac{S_3}3\Bigr\rfloor, \ldots , \Bigl\lfloor \frac{S_N}N\Bigr\rfloor$ stones. So you can pretend that you are playing Nim, and follow the usual Nim strategy.
For example, suppose we have piles $S_1 = 3, S_2 = 4, S_3 = 7$. We can pretend this is a Nim game with piles 3, 2, and 2, and play accordingly. In this case there are several winning Nim moves. One such is to take 1 stone from the third pile; this corresponds to taking 3 stones from $S_3$, moving to $(3, 4, 4)$.
Addendum Sorry, I misread the question and thought a legal move was to remove any multiple of $i$ stones from pile $S_i$. But the game as given is even simpler. Again we pretend that instead of $S_1,\ldots,S_N$ the piles contain $S_1, \Bigl\lfloor \frac{S_2}2\Bigr\rfloor,\Bigl\lfloor \frac{S_3}3\Bigr\rfloor, \ldots , \Bigl\lfloor \frac{S_N}N\Bigr\rfloor$, and the rule is that a legal move removes one stone from any pile. It is now evident that a position is a win for the previous player if and only if $$\sum{\biggl\lfloor \frac{S_i}{i} \biggr\rfloor}$$
is even, and that it doesn't even matter what moves are made.