2020 AMC 10A Problem 11

For a number near the median, the sorted list includes all ordinary integers up to that number and all squares up to that number. Since $44^2=1936$ and $45^2=2025$, there are $44$ squares not exceeding any number from $1936$ through $2020$. At $1975$, there are $1975+44=2019$ list entries at most $1975$. At $1976$, there are $2020$ entries at most $1976$, so the $2020$th entry is $1976$, and the next is $1977$. The median is $1976.5$. Thus, C is the correct answer.