33. Find the minimum value of $\sum_{k=1}^{100}|n-k|$, where $n$ ranges over all positive integers.
33. Answer: 2500
It is given that $n$ is a positive integer.
For $n \geq 100, \sum_{k=1}^{100}|n-k|=100 n-5050$, so that its minimum value is 4950 which occurs at $n=100$.
For $n<100, \sum_{k=1}^{100}|n-k|=n^{2}-101 n+5050$. If $n$ is a positive integer, its minimum value can occur at either $n=50$ or $n=51$ only. By direct checking, its minimum value is 2500 .