Solution:
Let $r_{1}, r_{2}, \cdots, r_{2016}$ be the zeroes of $f(x)$. We can then write $f$ as
$$f(x) = c\left(x - r_{1}\right)\left(x - r_{2}\right) \cdots \left(x - r_{2016}\right)$$
where $\sum_{i=1}^{2016} r_{i} = S$. Thus
$$f(2x-3) = c\left(2x-3 - r_{1}\right)\left(2x-3 - r_{2}\right) \cdots \left(2x-3 - r_{2016}\right)$$
which has zeroes
$$\frac{r_{1} + 3}{2}, \frac{r_{2} + 3}{2}, \ldots, \frac{r_{2016} + 3}{2}$$
This means that the required sum is
$$\sum_{i=1}^{2016} \frac{r_{i} + 3}{2} = \frac{1}{2}\left[\sum_{i=1}^{2016} r_{i} + 3 \times 2016\right] = \frac{1}{2} S + 3024$$