My solution We claim that the answer is $M=\frac{1}{1-c}$ We know that $\sum_{k=1}^{m}(cn-k)a_{k}=\sum_{k=m+1}^{n}(k-cn)a_{k}$ because $c\in\left(\frac{1}{2},1\right)$ and $m=[cn]$ so $n\ge{m+1}>c\ge{m}\ge{1}$ notice that $0<a_1\le a_2\le\ldots \le a_m$ and $0\le cn-m\le cn-m+1\le\ldots \le cn-1$ so applying chebyshev inquailty we obtain $\sum_{k=1}^{m}(cn-k)a_{k}\le{\frac{1}{m}\sum_{k=1}^{m}(cn-k)\sum_{k=1}^{m}a_{k}}\le{(cn-\frac{m+1}{2})\sum_{k=1}^{m}a_{k}}$ also $0<a_{m+1}\le a_{m+2}\le\ldots \le a_n$ and $0<m+1-cn\le\ldots \le n-cn$ $(\frac{m+n+1}{2}-cn)\sum_{k=m+1}^{n}a_k=\frac{1}{n-m}\sum_{k=m+1}^{n}(k-cn)\sum_{k=m+1}^{n}a_k\le{\sum_{k=m+1}^{n}(k-cn)a_k}$ so we get the most important inquality $(\frac{m+n+1}{2}-cn)\sum_{k=m+1}^{n}\le{(cn-\frac{m+1}{2})\sum_{k=1}^{m}a_{k}}$ then we have $\sum_{k=1}^{n}a_k\le{\frac{n}{m+n+1-2cn}\sum_{k=1}^{m}}=\frac{1}{1+\frac{m+1}{n}-2c}\sum_{k=1}^{m}a_k$ but $m+1>cn$ hence $\frac{1}{1+\frac{m+1}{n}-2c}<\frac{1}{1-c}$ On the other hand if $n\ge \frac{1}{2c-1}$we can take $a_1=a_2=\ldots=a_m=1$ and $a_{m+1}=a_{m+2}=\ldots=a_n=\frac{2cnm-m(m+1)}{(n-m)(n+m+1)-2cn(n-m)}$ so $a_{m+1}\ge 1$ and satisfied $\frac{1}{n}\sum_{k=1}^{n}ka_{k}=c\sum_{k=1}^{n}a_{k}$ so $M\ge \frac{1}{1-c}-\frac{1}{n(1-c)^2}$ when n is sufficently large we get $M\ge \frac{1}{1-c}$ We are done.