2021 AMC 10A Problem 16

The list contains $$\begin{aligned} 1+2+\cdots+200&=\frac{200\cdot201}{2}\\ &=20100 \end{aligned}$$ entries, so its two middle positions are $10050$ and $10051.$ There are $$\frac{141\cdot142}{2}=10011$$ entries through the last $141,$ and $$\frac{142\cdot143}{2}=10153$$ entries through the last $142.$ Thus both middle entries are $142,$ so the median is $142.$ Thus, C is the correct answer.