Note that if $ r $ is a root, then so is $ -r, $ so the roots are of the form $ p, $ $ -p, $ $ q, $ $ -q, $ for some complex numbers $ p $ and $ q $. Since there are only two distinct roots, at least two of these values must be equal. If $ p = -p, $ then $ p = 0 $ is a root. Hence, setting $ x = 0, $ we must get 0. In other words, $ a^3 = 0, $ so $ a = 0 $. But then the polynomial is \[x^4 - x^2 = x^2 (x - 1)(x + 1) = 0,\]so there are three roots. Hence, there are no solutions in this case. Otherwise, $ p = \pm q, $ so the roots are of the form $ p, $ $ p, $ $ -p, $ $ -p, $ and the quartic is \[(x - p)^2 (x + p)^2 = x^4 - 2p^2 x^2 + p^4.\]Matching coefficients, we get $ -2p^2 = a^2 - 1 $ and $ p^4 = a^3 $. Then $ p^2 = \frac{1 - a^2}{2}, $ so \[\left( \frac{1 - a^2}{2} \right)^2 = a^3.\]This simplifies to $ a^4 - 4a^3 - 2a^2 + 1 = 0 $. Let $ f(x) = x^4 - 4x^3 - 2x^2 + 1 $. Since $ f(0.51) > 0 $ and $ f(0.52) < 0, $ there is one root in the interval $ (0.51,0.52) $. Since $ f(4.43) < 0 $ and $ f(4.44) > 0, $ there is another root in the interval $ (4.43,4.44) $. Factoring out these roots, we are left with a quadratic whose coefficients are approximately \[x^2 + 0.95x + 0.44 = 0.\]The discriminant is negative, so this quadratic has two distinct, nonreal complex roots. Therefore, all the roots of $ a^4 - 4a^3 - 2a^2 + 1 = 0 $ are distinct, and by Vieta's formulas, their sum is $ \boxed{4} $.