The value of $\underset{n\to \infty }{\lim }\left[\frac{1}{n^2}\sum _{1\le i<j\le n}{\tan }^{-1}\left(\frac{i}{n}\right){\tan }^{-1}\left(\frac{j}{n}\right)\right](i,j\in integers)$ is 

$$\begin{gathered} L=\underset{n\to \infty }{\lim }\left[\frac{1}{n^2}\sum _{1\le i<j\le n}{\tan }^{-1}\left(\frac{i}{n}\right){\tan }^{-1}\left(\frac{i}{n}\right)\right] \\ \Rightarrow \underset{n\to \infty }{\lim }\frac{1}{2n^2}[{\left(\sum _{i=1}^n{\tan }^{-1}\left(\frac{i}{n}\right)\right)}^2-\left(\sum _{i=1}^n{\left({\tan }^{-1}\left(\frac{i}{n}\right)\right)}^2\right)] \\ =\frac{1}{2}{\left({\int }_0^1{\tan }^{-1}\left(x\right)dx\right)}^2-\frac{1}{2n}{\int }_0^1\left({\left({\tan }^{-1}\left(x\right)\right)}^2\right)dx \end{gathered}$$

Since $n\to \infty$, we obtain:

$L=\frac{1}{2}{\left({\int }_0^1{\tan }^{-1}\left(x\right)dx\right)}^2=\frac{1}{2}{\left(\frac{\pi -\ln \left(4\right)}{4}\right)}^2$