Let $X$ be a set of $n$ numbers with the mean $m$ and the SD $\sigma$. Let the maximum deviation $$D=max\{|x-m|:x\in X\}$$. Then $$D^2/n\leq\sigma^2$$ so $$D\leq\sqrt{n}\sigma$$ is an upper bound of $D$. Equality doesn't hold for $n>1$, because $$\sum_{x\in X}(x-m)=0$$ and I would like to find a lesser upper bound, or ideally, the supremum for $D$ of $n$ numbers, say $D_n$. Clearly $D_1=0$ and for $x_1\leq x_2$ $$D_2=\frac{x_2-x_1}{2}=\frac{1}{\sqrt{2}}\sigma$$, which are the exact deviations.
For $n\geq3$, I have considered the bipartite set $$X=\{m+d,m-\frac{d}{n-1},m-\frac{d}{n-1},\dots\}$$ with $d>0$, for which $d=\sqrt{n-1}\sigma$. But I'm not sure if $$D_n\leq\sqrt{n-1}\sigma$$. Any suggestions or known facts?