This is not an answer, but since no one has offered anything better, I thought that it might be worthwhile to make a compilation of results for small numbers of piles. It’s complete for fewer than five piles and makes a start on the case of five piles.
Clearly Alice wins when there is one pile by taking the whole pile.
If there are two piles, with $m$ and $n$ stones respectively, Alice wins when $m\ne n$: she takes $|m-n|$ from the larger pile, and no matter what Bob then does, he must either leave her a single pile or two piles of unequal size.
Alice wins when there are three piles. If two are equal in size, she takes all of the third pile. Otherwise, if the piles are of sizes $a<b<c$, she takes $c-(b-a)$ from $c$ and transfers $b-a$ to $a$, leaving Bob two piles of size $b$.
Any initial position of the form $\langle m,m,n,n\rangle$, where $m,n>0$, is a win for Bob: if Alice takes a whole pile, Bob wins as first player from the resulting three-pile game, and if Alice does anything else, Bob can leave her another position of this type. It follows that any initial position of the form $\langle k,k,m,n\rangle$ with $0<k\le m<n$ is a win for Alice, since she can leave Bob a position of the first type. This leaves only positions of the form $\langle k,\ell,m,n\rangle$ with $0<k<\ell<m<n\rangle$. The simplest of these is $\langle 1,2,3,4\rangle$, which is a win for Alice, as she can leave $\langle 1,1,3,3\rangle$ by taking $2$ from the fourth pile and transferring one to the second pile. In general Alice can take $$n-(m-\ell)-k=(n-m)+(\ell-k)>0$$ from the fourth pile and transfer $m-\ell$ to the second, leaving $\langle k,k,m,m\rangle$, so Alice can win from these positions. In short, Alice wins unless the initial position is of the form $\langle m,m,n,n\rangle$ with $0<m\le n$.
With five piles Alice wins if there are two piles of the same size by reducing the other three piles to a pair of piles of the same size. Thus, we need only consider the case in which all five piles are of different sizes. Bob will win if the initial position is $\langle 1,2,3,4,5\rangle$: Alice must leave him either four piles, or five piles, two of which are the same size. Otherwise, the player who first leaves the other player the position $\langle 1,2,3,4,5\rangle$ will win. This implies, for instance, that a five-pile position that differs from $\langle 1,2,3,4,5\rangle$ in exactly one position is a win for Alice: either it contains two piles of the same size, or she can reduce it to $\langle 1,2,3,4,5\rangle$.
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
This is known as Wythoff's game.
The losing positions are $(\lfloor n \varphi\rfloor, \lfloor n \varphi^2\rfloor)$ where $\varphi$ is the golden ratio.
An interesting way of presenting this game is to have the players move a Queen on a chessboard, which you can find at cut-the-knot here.