If the standard deviation of the data 1, 3, 5,7, ... , 2017 is $\sigma$ then $\left[\sigma \right]$ is (where $\left[.\right]$ denotes the greatest interger function)

Let $\overline{X}$ be the mean of the given data. Then,

$\overline{X}=\frac{1+3+5+7+\dots +2017}{109}=\frac{(109{)}^2}{109}=109$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{Variance }\left({\sigma }^2\right)=\frac{1}{1009}\left(1^2+3^2+\dots +{2017}^2\right)-(1009{)}^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\sigma }^2=339360\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sigma =582.54\end{array}$

Hence, $\left[\sigma \right]=582$