jmc

By the formula for an infinite geometric series, $$S(r)=\frac{12}{1-r}.$$Thus, we are given that $$S(a)S(-a)=\frac{12}{1-a}\cdot \frac{12}{1+a}=\frac{144}{1-a^2}=2016.$$Instead of solving for $a$ explicitly, we note that $$\begin{array}{rl}S(a)+S(-a)& =\frac{12}{1-a}+\frac{12}{1+a}\\ & =\frac{12(1-a)+12(1+a)}{1-a^2}\\ & =\frac{24}{1-a^2}\\ & =\frac{1}{6}\cdot \frac{144}{1-a^2}\\ & =\frac{1}{6}\cdot 2016\\ & =\boxed{336},\end{array}$$which is the answer.