Suppose $N> 2016$. Then Bob can simply play $1,-1,1,-1,\dots$, and Ana can achieve at most $2016$. So $N\le 2016$.

I believe any even number from 0 to 2016 works.

Bumplease

Answer $N$ is any even number from $0$ to $2016$. Proof It is necessary for $N$ to be an even number from $0$ to $2016$. If Bob chooses the sequence $(1, -1, 1, -1, \dots, 1, -1)$, then the most Ana can get is $2016$ by making the blocks $((1), (-1), (1), (-1), \dots, (1), (-1))$. Also, the sum of any block Ana can make is $-1, 0,$ or $1$. Whenever Ana makes a block that sums to $-1$ or $1$, it takes an odd number of values. Whenever the block sums to $0$, there are an even number of values. So, there are an even number of blocks that sum to either $-1$ or $1$, which implies that $N$ must be even. It is sufficient for $N$ to be an even number from $0$ to $2016$. Now we show that all even values of $N$ work between $0$ and $2016$. First, suppose Ana has a sequence of $N\ge 4$ blocks, such that the sum in each block is either $-1$ or $1$. We show that it is possible to find another sequence with $N-2$ blocks all summing to either $-1$ or $1$. Let the blocks be $$B_N = (b_1, b_2, \dots, b_N)$$There exists $2$ consecutive blocks, $b_i$ and $b_{i+1}$ that sum to $0$. Otherwise, all $b_j$ are of the same sign, which implies that $$|b_1 + b_2 + \cdots + b_N| > 0,$$impossible as $$b_1 + b_2 + \cdots + b_N = 1 + (-1) + 1 + (-1) + \cdots + (-1) = 0.$$If $i=0$, take $b_{0}, b_{1}, b_{2}$ and combine them into one block. Otherwise, take $b_{i-1}, b_i, b_{i+1}$ and combine them into one block. We have found a new sequence of $N-2$ blocks that each sum to either $-1$ or $1$. This shows all even numbers from $2$ to $2016$ are possible. What about $0$? That can be achieved by a single block with all of the numbers. $\blacksquare$