If we want the final value to be as large as possible, we want $x_2,x_3,...,x_100$ to be as small as possible. So, we have $|x_2|=64$. Since we want $x_2$ to be as small as possible, we make it $-64$. We keep on doing this for all of the other numbers. So, our numbers are $63,-64,-65,-66,...,-162$. Therefore, our value is $-63+64+65+...+162=-63+\frac{64+162}{2}\cdot 99=11187-63=\boxed{11124}$, unless I did something wrong.

To remove the absolute values, we square the given equation, which gives us $x_{n + 1}^2 = x_n^2 + 2x_n + 1$. We now have $x_2^2 = x_1^2 + 2x_1 + 1$ $x_3^2 = x_2^2 + 2x_2 + 1$ ... $x_{100}^2 = x_{99}^2 + 2x_{99} + 1$. Summing these up, we have $x_{100}^2 = x_1^2 + 2(x_1 + x_2 + ... + x_{99}) + 99 \implies x_1 + x_2 + ... + x_{99} = \dfrac{x_{100}^2 - x_1^2 - 99}{2}$ $= \dfrac{x_{100}^2 - 63^2 - 99}{2} = \dfrac{x_{100}^2 - 4068}{2}$. It now suffices to minimize $x_1 + x_2 + ... + x_{100} = x_{100} + \dfrac{x_{100}^2 - 4068}{2} = \dfrac{1}{2}\left(x_{100}^2 + 2x_{100} - 4068\right)$ $= \dfrac{1}{2}\left((x_{100} + 1)^2 - 4069)\right) \le -\dfrac{4069}{2}$. However, it is clear that all of the odd terms of the sequence will have odd values, and all of the even terms of the sequence will have even values. Therefore the minimum value of $(x_{100} + 1)^2$ occurs at $x_{100} = 0$ (or $-2$), in which case we have $x_1 + x_2 + ... + x_{100} = \dfrac{1}{2}\left(0 + 1)^2 - 4069)\right) = -2034$. Then $-x_1 - x_2 + ... - x_{100} = 2034$.