Median of the first 2023 integers and their squares

Form a list containing the integers $1,2,\dots,2023$ together with the perfect squares $1^2,2^2,\dots,2023^2$ (so $4046$ entries, duplicates kept). Find the median of this list.

$1979.5$. The median is the average of the 2023rd and 2024th smallest entries. Entries at most $1979$ are the $1979$ integers plus the $44$ squares up to $44^2=1936$, i.e. exactly $2023$ entries; the next entry is $1980$. So the median is $\tfrac{1979+1980}{2}$.