1. Let $n$ be a positive integer, then the minimum value of $|n-1|+|n-2|+\cdots \cdots+|n-100|$ is ).
A. 2500
B. 4950
C. 5050
D. 5150
1. A

When $n \geqslant 100$, the sum mentioned is $100 n-5050$. It reaches its minimum value of 4950 when $n=100$. When $1 \leqslant n \leqslant 100$, the sum is $(n-1)+\cdots+(n-n)+(n+1-n)+\cdots+(100 \quad n)=n^{2}-101 n+5050=\left(n-\frac{101}{2}\right)^{2}-\left(\frac{101}{2}\right)^{2}+5050$.
Therefore, the minimum value of 2500 is reached when $n=50$ or 51.