What is the median of the following list of $4040$ numbers$?$
\[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\]
$\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$
We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$. Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$. 
Since there are $44$ duplicates/extras, it will shift up our median's placement down $44$. Had the list of numbers been $1,2,3, \dots, 4040$, the median of the whole set would be $\dfrac{1+4040}{2}=2020.5$. 
Thus, our answer is $2020.5-44=\boxed{\textbf{(C)}\ 1976.5}$.
~aryam
~Additions by BakedPotato66