Inversion of Double Summations.

$\begin{aligned} S & = \sum_{\color{#3D99F6}\beta=1}^{20} \sum_{\color{#D61F06}\alpha=1}^{20} \tan^{-1} \left(\frac {\color{#D61F06}\alpha}{\color{#3D99F6}\beta} \right) & \small \color{#3D99F6} \text{Add another }S \text{ with }\alpha \text{ and }\beta \text{ swapped.} \\ & = \frac 12 \left(\sum_{\color{#3D99F6}\beta=1}^{20} \sum_{\color{#D61F06}\alpha=1}^{20} \tan^{-1} \left(\frac {\color{#D61F06}\alpha}{\color{#3D99F6}\beta} \right) + \sum_{\color{#D61F06}\alpha=1}^{20} \sum_{\color{#3D99F6}\beta=1}^{20} \tan^{-1} \left(\frac {\color{#3D99F6}\beta}{\color{#D61F06}\alpha} \right) \right) & \small \color{#3D99F6} \text{Swapped and order of summation.} \\ & = \frac 12 \left(\sum_{\color{#3D99F6}\beta=1}^{20} \sum_{\color{#D61F06}\alpha=1}^{20} \tan^{-1} \left(\frac {\color{#D61F06}\alpha}{\color{#3D99F6}\beta} \right) + \sum_{\color{#3D99F6}\beta=1}^{20} \sum_{\color{#D61F06}\alpha=1}^{20} \tan^{-1} \left(\frac {\color{#3D99F6}\beta}{\color{#D61F06}\alpha} \right) \right) \\ & = \frac 12 \sum_{\beta=1}^{20} \sum_{\alpha=1}^{20} \left(\tan^{-1} \left(\frac {\alpha}{\beta} \right) + \tan^{-1} \left(\frac {\beta}{\alpha} \right) \right) \\ & = \frac 12 \sum_{\beta=1}^{20} \sum_{\alpha=1}^{20} \left(\tan^{-1} \left(\frac {\alpha}{\beta} \right) + \cot^{-1} \left(\frac \alpha \beta \right) \right) \\ & = \frac 12 \sum_{\beta=1}^{20} \sum_{\alpha=1}^{20} \left(\tan^{-1} \left(\frac {\alpha}{\beta} \right) + \frac \pi 2 - \tan^{-1} \left(\frac \alpha \beta \right) \right) \\ & = \frac 12 \sum_{\beta=1}^{20} \sum_{\alpha=1}^{20} \frac \pi 2 = \frac {20\times 20}2 \times \frac \pi 2 = 100 \pi \end{aligned}$

Therefore, $\dfrac S\pi = \boxed{100}$ .

Important Identity for this problem:

$\tan ^{-1} \frac {a}{b} + \tan ^{-1} \frac {b}{a}=\frac {π}{2}$

(I have denoted $\alpha$ by $a$ and $\beta$ by $b$ in this solution)

There will be a total of 20×20=400 terms

For every $\frac {a}{b}$ term there will be a $\frac {b}{a}$ term, except when $a=b$ . Addition of those two terms will give us π/2 for every pair.

There are 20 terms where $a=b$ , and 380 other terms. Thus, the total number of pairs that will give $\frac {π}{2}$ is $\frac {380}{2}=190$ .

The other 20 terms will each have $a=b$ , and $\tan ^{-1} 1=\frac {π}{4}$

Therefore, the answer is $190\cdot \frac {π}{2} + 20\cdot \frac {π}{4}=100π$