It's all about number theory! (3)

the average $\overline{x}$ where minimal while sum of the elements of the list where minimal, also i can't have the same number 2 times in the list, then the $\overline{x}$ can have here minimun value whene $\overline{x}=\frac{2015+\sum_{k=0}^{n}{k}}{n+1}$ $min\{\overline{x}\}=min\{\frac{2015+\sum_{k=0}^{n}{k}}{n+1}\}$ $min\{\overline{x}\}=min\{\frac{2015}{n+1}+\frac{1}{2}n\}$ $min\{\overline{x}\}\rightarrow\frac{d}{dn}(\frac{2015}{n+1}+\frac{1}{2}n)=0$ $min\{\overline{x}\}\rightarrow\frac{1}{2}=\frac{2015}{(n+1)^2}$ $min\{\overline{x}\}\rightarrow n=\left\lfloor{\sqrt{2015*2}}-1\right\rfloor$ $min\{\overline{x}\}\rightarrow n=62$