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, we compute $f(x)$ for any $x$ such that $\frac{2-3x}{x+3}+x^2$ is an integer. If the corresponding integer is $n,$ we have
\[n \le \frac{2-3x}{x+3}+x^2 < n+1.\]Multiplying both sides by $x+3,$ we get
\[n(x+3) \le 2-3x+x^2(x+3) < (n+1)(x+3).\]Expanding, we get
\[nx + 3n \le 2-3x+ x^3+3x^2 < nx + 3n + 3 + 3x.\]Then
\[mx + 3m \le 2 + x^3 + 3x^2 - 3x + 3n < mx + 3m + 3 + 3x\]for some integer $m.$ Since $m$ is simply $n$ divided by $x+3,$ we want $n$ to be an integer multiple of $x+3.$
Note that $x+3$ divides $n$ if and only if $x+3$ divides $2 + x^3 + 3x^2 - 3x + 3n - 3(x+3),$ or
\[x+3 \mid -x^3 - 3x^2 + 3n - 7.\]Let $g(x) = -x^3 - 3x^2.$ Then $g'(x) = -3x^2 - 6x.$ Suppose $-3 \le x \le 1.$ Then $g'(x) \le 3$ and $g(1) = -4,$ so
\[g(x) \le -4.\]If $x \ge 4,$ then $g'(x) \ge 36 - 18 = 18,$ so $g(x) \ge -108.$ Hence, for $-3 \le x,$
\[|g(x)| \le 108.\]So, $|g(x) + 3n - 7| \le 108 + 3n + 7,$ or
\[n - \frac{115}{3} \le x + \frac{7}{3n} = d \le n + \frac{115}{3}.\]Thus, $d - n \le \frac{115}{3}.$ This means that
\[2 - 3d + d^3 + 3d^2 + 3n < 6d + 6x + 9 < 70,\]provided that $d \le 333.$ Thus, the possible values of $d$ for $n = 1$ are 1, 4, 7, $\dots,$ $331.$ The possible values of $d$ for $n = 2$ are 2, 5, 8, $\dots,$ $332,$ and so on.
Since there are 333 numbers for $n = 1,$ 333 numbers for $n = 2,$ and so on, the sum of all possible values of $f(x)$ over the range $1 \le x \le 1000$ is
\[(333 \cdot 1 + 333 \cdot 2 + 333 \cdot 3 + \dots + 333 \cdot 333) + 334 \cdot 333 = 334 \cdot 333 \cdot \frac{333}{2} + 334 \cdot 333 = \boxed{334 \cdot 333 \cdot 334}.\]