Let f(x) = \left \lfloor \frac{2 - 3x}{x + 3} + x^2 \right \rfloor.
Find f(1) + f(2) + f(3) + ... + f(999) + f(1000).
First, note that $\frac{2-3x}{x+3} = -3 + \frac{11}{x+3}.$ Also note that as $x$ approaches infinity, the expression approaches $-3.$
For values of $x$ from $1$ to $333,$ the expression $-3 + \frac{11}{x+3}$ takes on values ranging from $8$ to $-3.$ Therefore, for these values of $x,$ $f(x) = 8.$
As for $x = 334,$ we have $-3 + \frac{11}{x+3} = 7 \frac{5}{337}.$ Therefore, $f(334) = 7.$
Likewise, for $x$ from $335$ to $998,$ we have $f(x) = 7$ since $-3 + \frac{11}{x+3}$ is less than $7 \frac{5}{337},$ but greater than $7.$
Finally, for $x = 999$ and $x = 1000,$ we have $-3 + \frac{11}{x+3}$ less than $6\frac{5}{998},$ but greater than $6.$ Therefore, $f(999) = f(1000) = 6.$
Putting the pieces together,
\[f(1) + f(2) + f(3) + \dots + f(999) + f(1000) = (333)(8) + (7)(664) + (6)(2) = \boxed{4394}.\]