In order to factor the polynomial, we will try to solve the equation $ x^8 + 98x^4 + 1 = 0 $. First, we can divide both sides by $ x^4, $ to get $ x^4 + 98 + \frac{1}{x^4} = 0, $ so \[x^4 + \frac{1}{x^4} = -98.\]Then \[x^4 + 2 + \frac{1}{x^4} = -96,\]which we can write as $ \left( x^2 + \frac{1}{x^2} \right)^2 = -96 $. Hence, \[x^2 + \frac{1}{x^2} = \pm 4i \sqrt{6}.\]Then \[x^2 - 2 + \frac{1}{x^2} = -2 \pm 4i \sqrt{6},\]which we can write as \[\left( x - \frac{1}{x} \right)^2 = -2 \pm 4i \sqrt{6}.\]To work with this equation, we will find the square roots of $ -2 \pm 4i \sqrt{6} $. Assume that $ \sqrt{-2 + 4i \sqrt{6}} $ is of the form $ a + b $. Squaring, we get \[-2 + 4i \sqrt{6} = a^2 + 2ab + b^2.\]We set $ a^2 + b^2 = -2 $ and $ 2ab = 4i \sqrt{6}, $ so $ ab = 2i \sqrt{6} $. Then $ a^2 b^2 = -24, $ so $ a^2 $ and $ b^2 $ are the roots of the quadratic \[t^2 + 2t - 24 = 0,\]which factors as $ (t - 4)(t + 6) = 0 $. Hence, $ a^2 $ and $ b^2 $ are 4 and $ -6 $ in some order, which means $ a $ and $ b $ are $ \pm 2 $ and $ \pm i \sqrt{6} $ in some order. We can check that \[(2 + i \sqrt{6})^2 = 4 + 4i \sqrt{6} - 6 = -2 + 4i \sqrt{6}.\]Similarly, \begin{align*} (-2 - i \sqrt{6})^2 &= -2 + 4i \sqrt{6}, \\ (2 - i \sqrt{6})^2 &= -2 - 4i \sqrt{6}, \\ (-2 + i \sqrt{6})^2 &= -2 - 4i \sqrt{6}.\end{align*}Thus, \[x - \frac{1}{x} = \pm 2 \pm i \sqrt{6}.\]If \[x - \frac{1}{x} = 2 + i \sqrt{6},\]then \[x - \frac{1}{x} - 2 = i \sqrt{6}.\]Squaring both sides, we get \[x^2 - 4x + 2 + \frac{4}{x} + \frac{1}{x^2} = -6,\]so \[x^2 - 4x + 8 + \frac{4}{x} + \frac{1}{x^2} = 0.\]This simplifies to $ x^4 - 4x^3 + 8x^2 + 4x + 1 $. Similarly, \[x - \frac{1}{x} = -2 + i \sqrt{6}\]leads to $ x^4 + 4x^3 + 8x^2 - 4x + 1 $. Thus, \[x^8 + 98x^4 + 1 = (x^4 + 4x^3 + 8x^2 - 4x + 1)(x^4 - 4x^3 + 8x^2 + 4x + 1).\]Evaluating each factor at $ x = 1, $ the final answer is $ (1 + 4 + 8 - 4 + 1) + (1 - 4 + 8 + 4 + 1) = \boxed{20} $.