We want to know the $2020$-th term and the $2021$-st term to get the median.  We know that $44^{2}=1936$. So, numbers $1^{2}, 2^{2}, \ldots, 44^{2}$ are in between $1$ and $1936$.  So, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$-th number.  Also, notice that $45^{2}=2025$, which is larger than $2021$.  Then the $2020$-th term will be $1936+40=1976$, and similarly the $2021$-th term will be $1977$.  Solving for the median of the two numbers, we get (C) $1976.5$. 
What is the median of the following list of 4040 numbers: $$ 1,2,3, \ldots, 2020,1^{2}, 2^{2}, 3^{2}, \ldots, 2020^{2}$$? (2020 AMC 10A Problems, Question \#11) 
Options:
[a] $$1974.5$$ 
[b] $$1975.5$$ 
[c] $$1976.5$$ 
[d] $$1977.5$$ 
[e] $$1978.5$$ 