What is greatest integer function?

$\begin{aligned} S & = \lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \lfloor \sqrt{4} \rfloor + \lfloor \sqrt{5} \rfloor + \lfloor \sqrt{6} \rfloor + \cdots + \lfloor \sqrt{2004} \rfloor \\ & = \underbrace{1}_{1^\text{st} \text{ term}}+1+1+\underbrace{2}_{4^\text{th} \text{ term}}+2+2+...+\underbrace{3}_{9^\text{th} \text{ term}}+3+3...+44 \\ & = \sum_{n=1}^{43} ((n+1)^2-n^2)n + (2004-44^2+1)(44) \\ & = \sum_{n=1}^{43} (2n+1)n + 3036 \\ & = 2 \sum_{n=1}^{43} n^2 + \sum_{n=1}^{43} n + 3036 \\ & = 2\left(\frac{43\cdot 44 \cdot 87}6 \right) + \frac {43 \cdot 44}2 + 3036 \\ & = 54868 + 946 + 3036 \\ & = \boxed{58850} \end{aligned}$

There are 3, '1's,...5,'2's....7,'3's.... . . . (2n+1), 'n's.
$\sqrt{2004}=44.76605857,\ \ so \ above \ n=43,\ +2004-(44^2)+1=69\ '44's.\\ \displaystyle \sum_{n=1}^{43}(2n+1)*n+69*44=\sum_{n=1}^{43}(2n^2+n)+69*44=2*\dfrac{n*(n+1)(2n+1)} 6 +\dfrac{n(n+1)} 2|_{n_{43}} +69*44\\ =\frac 1 3 *43*44*87+ \frac 1 2*43*44+69*44=\huge\ \ \ \color{#D61F06}{58850}$

(1×3)+(2×5)+(3×7)+(4×9)+...=sum (n×(2n+1)) till n=43 and then 69×44, because with 44 only 69 numbers are there in the question