To evaluate the sum $f(1) + f(2) + f(3) + \dots + f(999) + f(1000)$ where

$f(x) = \left\lfloor \frac{3x - 5}{x + 4} \right\rfloor,$

we start by simplifying the expression inside the floor function:

$\frac{3x - 5}{x + 4}.$

Let's perform polynomial long division to express this fraction as a sum of a linear term and a fractional term:

1. Divide $3x$ by $x$ to get 3.
2. Multiply $3$ by $x + 4$ to get $3x + 12$.
3. Subtract $3x + 12$ from $3x - 5$:

$3x - 5 - (3x + 12) = -17.$

Thus,

$\frac{3x - 5}{x + 4} = 3 - \frac{17}{x + 4}.$

So,

$f(x) = \left\lfloor 3 - \frac{17}{x + 4} \right\rfloor.$

To find the value of $f(x)$, notice that $3 - \frac{17}{x + 4}$ is a number slightly less than 3, because $\frac{17}{x + 4}$ is positive. Therefore,

$f(x) = \left\lfloor 3 - \frac{17}{x + 4} \right\rfloor = 2.$

$f(x) = 2$ holds as long as

$3 - \frac{17}{x + 4} \geq 2,$

which simplifies to:

$-\frac{17}{x + 4} \geq -1 \quad \text{or} \quad \frac{17}{x + 4} \leq 1.$

Cross-multiplying gives:

$17 \leq x + 4 \quad \text{or} \quad x \geq 13.$

Thus, for $x \geq 13$, $f(x) = 2$.

Next, let's check the values of $f(x)$ for $x < 13$:

• For $x = 1$, $f(1) = \left\lfloor \frac{-2}{5} \right\rfloor = -1$.
• For $x = 2$, $f(2) = \left\lfloor \frac{1}{6} \right\rfloor = 0$.
• For $x = 3$, $f(3) = \left\lfloor \frac{4}{7} \right\rfloor = 0$.
• For $x = 4$, $f(4) = \left\lfloor \frac{7}{8} \right\rfloor = 0$.
• For $x = 5$, $f(5) = \left\lfloor \frac{10}{9} \right\rfloor = 1$.
• For $x = 6$, $f(6) = \left\lfloor \frac{13}{10} \right\rfloor = 1$.
• For $x = 7$, $f(7) = \left\lfloor \frac{16}{11} \right\rfloor = 1$.
• For $x = 8$, $f(8) = \left\lfloor \frac{19}{12} \right\rfloor = 1$.
• For $x = 9$, $f(9) = \left\lfloor \frac{22}{13} \right\rfloor = 1$.
• For $x = 10$, $f(10) = \left\lfloor \frac{25}{14} \right\rfloor = 1$.
• For $x = 11$, $f(11) = \left\lfloor \frac{28}{15} \right\rfloor = 1$.
• For $x = 12$, $f(12) = \left\lfloor \frac{31}{16} \right\rfloor = 1$.

The results are:

• $f(1) = -1$
• $f(2) = f(3) = f(4) = 0$
• $f(5) = f(6) = f(7) = f(8) = f(9) = f(10) = f(11) = f(12) = 1$
• $f(x) = 2$ for $x \geq 13$.

The sum $f(1) + f(2) + \dots + f(1000)$ can now be computed by adding:

• The first term, $f(1) = -1$.
• The next three terms, $f(2) + f(3) + f(4) = 0$.
• The next eight terms, $f(5) + \dots + f(12) = 8$.
• The remaining $988$ terms (from $f(13)$ to $f(1000)$), each equal to 2: $2 \times 988 = 1976$.

Adding these gives:

$\text{Total sum} = -1 + 0 + 8 + 1976 = 1983.$

Thus, the final answer is:

$\boxed{1983}.$

Would you like more details or have any questions? Here are some related questions you might find useful:

1. What is the general process for polynomial long division?
2. How do floor functions behave with fractional inputs?
3. Why does $f(x)$ change at $x = 13$ in this problem?
4. What other types of problems involve floor functions?
5. How can we graph the function $f(x)$ in this problem?

Tip: Always double-check the boundary values in problems involving piecewise functions, as these can often reveal key changes in behavior.