First, we simplify the absolute values by considering two cases: $x \geq 1$ and $x \leq -1$.

Case 1: $x \geq 1$
The equation becomes $2x + 2y = x^2 - 1$, or equivalently, $x^2 - 2x - 2y - 1 = 0$. Using the quadratic formula, we get $$x = 1 + \sqrt{2y+2} \quad \text{or} \quad x = 1 - \sqrt{2y+2}.$$For each $y \geq 1$, there are two possible values of $x$, one positive and one negative. For $y = 0$, we get $x = \pm 1$, and for $y \leq -1$, there are no integral solutions. Therefore, there are $2 \cdot \lfloor \frac{n-1}{2} \rfloor + 2$ solutions in this case.

Case 2: $x \leq -1$
The equation becomes $-2x + 2y = x^2 - 1$, or equivalently, $x^2 + 2x - 2y - 1 = 0$. Using the quadratic formula, we get $$x = -1 - \sqrt{2y+2} \quad \text{or} \quad x = -1 + \sqrt{2y+2}.$$For each $y \geq 1$, there are two possible values of $x$, one positive and one negative. For $y = 0$, we get $x = -1$, and for $y \leq -1$, there are no integral solutions. Therefore, there are $2 \cdot \lfloor \frac{n-1}{2} \rfloor + 1$ solutions in this case.

Putting the two cases together, we get a total of $$2\left(2 \cdot \lfloor \frac{n-1}{2} \rfloor + 2\right) + 2\left(2 \cdot \lfloor \frac{n-1}{2} \rfloor + 1\right) = 8 \lfloor \frac{n-1}{2} \rfloor + 6$$solutions, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. Therefore, the answer is $\boxed{8 \lfloor \frac{n-1}{2} \rfloor + 6}$.