Given $c \in\left(\frac{1}{2}, 1\right)$. Find the smallest constant $M$, such that for any integer $n \geqslant 2$ and real numbers $0<a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, if $\frac{1}{n} \sum_{k=1}^{n} k a_{k}=c \sum_{k=1}^{n} a_{k}$, then $\sum_{k=1}^{n} a_{k} \leqslant M \sum_{k=1}^{m} a_{k}$, where $m=[c n]$ denotes the greatest integer not exceeding $c n$.
Solution: The required minimum constant $M=\frac{1}{1-c}$.
$$
\begin{array}{l}
\because m=[c n], \text { and } c \in\left(\frac{1}{2}, 1\right), \\
\therefore c n-1c-\frac{2}{n}>\cdots>c-\frac{m}{n} \geqslant 0,
\end{array}
$$

and when $00, a_{m+1}$ $=a_{m+2}=\cdots=a_{n}>0$, and it can satisfy the given condition $\frac{1}{n} \sum_{k=1}^{n} k a_{k}=c \sum_{k=1}^{n} a_{k}$, therefore, the equality in inequalities (2) and (3) holds. Thus, the equality in inequality (4) holds, and we have
$$
\sum_{k=m+1}^{n} a_{k}=\frac{c-\frac{m+1}{2 n}}{\frac{n+m+1}{2 n}-c} \cdot \sum_{k=1}^{m} a_{k} \text {. }
$$

From equations (6) and (7), we get
$$
\begin{array}{l}
M-1 \geqslant \frac{\sum_{k=m+1}^{n} a_{k}}{\sum_{k=1}^{m} a_{k}}=\frac{c-\frac{m+1}{2 n}}{\frac{n+m+1}{2 n}-c} \\
\underset{(\because m \leqslant c n)}{\frac{n+c n+1}{2 n}-c}=\frac{c-\frac{c n+1}{2 n}}{1-c+\frac{1}{n}}, \\
\end{array}
$$

Therefore, $M \geqslant \frac{1}{1-c+\frac{1}{n}}-$ holds for all $n \geqslant 2$.
Let $n \rightarrow+\infty$. Then
$$
M \geqslant \lim _{n \rightarrow+\infty} \frac{1}{1-c+\frac{1}{n}}=\frac{1}{1-c} .
$$