Problem to tackle with

$\begin{aligned} S & = \sum_{r=1}^{16} \frac {8r}{4r^4+1} \\ & = \frac {8(1)}{4(1^4)+1} + \frac {8(2)}{4(2^4)+1} + \color{#3D99F6} \sum_{r=3}^{16} \frac {8r}{4r^4+1} & \small \color{#3D99F6} \text{Note that } \sum_{r=3}^{16} \frac {8r}{4r^4+1} < \sum_{r=3}^{16} \frac {8r}{4r^4} = \sum_{r=3}^{16} \frac 2{r^3} \\ & < \frac 85 + \frac {16}{65} + \color{#3D99F6} 2 \sum_{r=3}^{16} \frac 1{r^3} & \small \color{#3D99F6} \text{and } \sum_{r=3}^{16} \frac 1{r^3} < \sum_{r=3}^\infty \frac 1{r^3} = \zeta (3) - 1 - \frac 18 \\ & < \frac 85 + \frac {16}{65} + \color{#3D99F6} 2 \left(\zeta (3) - 1 - \frac 18 \right) & \small \color{#3D99F6} \text{Riemann zeta function } \zeta (3) \approx 1.202057 \\ & \approx 2.0003 \end{aligned}$

Therefore, $S < \boxed 2$ .