2021 AMC 12A Problem 16

The list has $1 + 2 + \cdots + 200$ $= \dfrac{200\cdot 201}{2}$ $= 20100$ terms, so the median is the average of the $10050$th and $10051$st terms. The value $n$ occupies positions up to $\dfrac{n(n+1)}{2}.$ Since $\dfrac{141\cdot 142}{2} = 10011$ and $\dfrac{142\cdot 143}{2} = 10153,$ positions $10012$ through $10153$ all equal $142.$ Both middle positions fall in this block, so the median is $142.$ Thus, the correct answer is C.