For any positive integer $M$ and probabilities $p_j$ for $j=1,2,...,M$, which are arranged in ascending order, i.e. $p_1 < p_2 < ...<p_M $, one has $\sum\limits_{j=1}^{M}(j-1)p_j \geq \frac{M-1}{2}$.
For uniform distribution i.e. $p_1=p_2=\ldots=p_M=1/M$, the equality holds
$\sum\limits_{j=1}^{M}(j-1)p_j= \frac{1}{M}\sum\limits_{j=1}^{M}(j-1)= \frac{1}{M}\frac{(M-1)M}{2}= \frac{(M-1)}{2}$.
But, how to prove for any probability distribution?
My intuition is, by using Lagrange Multipliers, we can prove that minimum value occurs for uniform distribution.