Let $f_1(x) = \frac{2}{3}-\frac{3}{3x+1}$, and for $n \ge 2$, define $f_n(x) = f_1(f_{n-1} (x))$. The value of x that satisfies $f_{1001}(x) = x - 3$ can be expressed in the form $\frac{m}{n}$,
where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
1. **Define the function \( f_1(x) \):**
   \[
   f_1(x) = \frac{2}{3} - \frac{3}{3x+1}
   \]

2. **Calculate \( f_2(x) \):**
   \[
   f_2(x) = f_1(f_1(x))
   \]
   Substitute \( f_1(x) \) into itself:
   \[
   f_2(x) = \frac{2}{3} - \frac{3}{3\left(\frac{2}{3} - \frac{3}{3x+1}\right) + 1}
   \]
   Simplify the inner expression:
   \[
   3\left(\frac{2}{3} - \frac{3}{3x+1}\right) + 1 = 2 - \frac{9}{3x+1} + 1 = 3 - \frac{9}{3x+1}
   \]
   Thus,
   \[
   f_2(x) = \frac{2}{3} - \frac{3}{3 - \frac{9}{3x+1}}
   \]
   Simplify the denominator:
   \[
   3 - \frac{9}{3x+1} = \frac{3(3x+1) - 9}{3x+1} = \frac{9x - 6}{3x+1}
   \]
   Therefore,
   \[
   f_2(x) = \frac{2}{3} - \frac{3(3x+1)}{9x-6} = \frac{2}{3} - \frac{3x+1}{3x-2}
   \]
   Simplify further:
   \[
   f_2(x) = -\frac{1}{3} - \frac{3}{3x-2}
   \]

3. **Calculate \( f_3(x) \):**
   \[
   f_3(x) = f_1(f_2(x))
   \]
   Substitute \( f_2(x) \) into \( f_1 \):
   \[
   f_3(x) = \frac{2}{3} - \frac{3}{3\left(-\frac{1}{3} - \frac{3}{3x-2}\right) + 1}
   \]
   Simplify the inner expression:
   \[
   3\left(-\frac{1}{3} - \frac{3}{3x-2}\right) + 1 = -1 - \frac{9}{3x-2} + 1 = -\frac{9}{3x-2}
   \]
   Thus,
   \[
   f_3(x) = \frac{2}{3} + \frac{3x-2}{3}
   \]

4. **Calculate \( f_4(x) \):**
   \[
   f_4(x) = f_1(f_3(x))
   \]
   Substitute \( f_3(x) \) into \( f_1 \):
   \[
   f_4(x) = \frac{2}{3} - \frac{3}{3\left(\frac{2}{3} + \frac{3x-2}{3}\right) + 1}
   \]
   Simplify the inner expression:
   \[
   3\left(\frac{2}{3} + \frac{3x-2}{3}\right) + 1 = 2 + 3x - 2 + 1 = 3x + 1
   \]
   Thus,
   \[
   f_4(x) = \frac{2}{3} - \frac{3}{3x+1} = f_1(x)
   \]

5. **Identify the periodicity:**
   \[
   f_4(x) = f_1(x)
   \]
   Therefore, \( f_n(x) \) is periodic with period 3. Specifically,
   \[
   f_{1001}(x) = f_{1001 \mod 3}(x) = f_2(x)
   \]

6. **Solve for \( x \) in \( f_{1001}(x) = x - 3 \):**
   \[
   f_2(x) = x - 3
   \]
   Substitute \( f_2(x) \):
   \[
   -\frac{1}{3} - \frac{3}{3x-2} = x - 3
   \]
   Multiply through by \( 3(3x-2) \):
   \[
   - (3x-2) - 9 = 3x(3x-2) - 9(3x-2)
   \]
   Simplify:
   \[
   -3x + 2 - 9 = 9x^2 - 6x - 27x + 18
   \]
   Combine like terms:
   \[
   -3x - 7 = 9x^2 - 33x + 18
   \]
   Rearrange to form a quadratic equation:
   \[
   9x^2 - 30x + 25 = 0
   \]
   Solve the quadratic equation using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{30 \pm \sqrt{900 - 900}}{18} = \frac{30}{18} = \frac{5}{3}
   \]

The final answer is \( \boxed{8} \)