How many positive integers $n$ are there such that the equation $\left \lfloor \sqrt[3] {7n + 2} \right \rfloor = \left \lfloor \sqrt[3] {7n + 3} \right \rfloor $ does not hold?

$ 
\textbf{a)}\ 0
\qquad\textbf{b)}\ 1
\qquad\textbf{c)}\ 7
\qquad\textbf{d)}\ \text{Infinitely many}
\qquad\textbf{e)}\ \text{None of above}
$
To determine how many positive integers \( n \) exist such that the equation 
\[
\left\lfloor \sqrt[3]{7n + 2} \right\rfloor = \left\lfloor \sqrt[3]{7n + 3} \right\rfloor
\]
does not hold, we need to analyze the conditions under which the floor functions of the cube roots of \( 7n + 2 \) and \( 7n + 3 \) are equal.

1. **Understanding the Floor Function**:
   The floor function \( \left\lfloor x \right\rfloor \) returns the greatest integer less than or equal to \( x \). Therefore, for the equation to hold, the integer parts of \( \sqrt[3]{7n + 2} \) and \( \sqrt[3]{7n + 3} \) must be the same.

2. **Analyzing the Cube Roots**:
   Let \( k = \left\lfloor \sqrt[3]{7n + 2} \right\rfloor \). This implies:
   \[
   k \leq \sqrt[3]{7n + 2} < k + 1
   \]
   Cubing all parts of the inequality, we get:
   \[
   k^3 \leq 7n + 2 < (k + 1)^3
   \]
   Similarly, for \( \left\lfloor \sqrt[3]{7n + 3} \right\rfloor \), we have:
   \[
   k \leq \sqrt[3]{7n + 3} < k + 1
   \]
   Cubing all parts of this inequality, we get:
   \[
   k^3 \leq 7n + 3 < (k + 1)^3
   \]

3. **Combining the Inequalities**:
   For both inequalities to hold simultaneously, we need:
   \[
   k^3 \leq 7n + 2 < (k + 1)^3 \quad \text{and} \quad k^3 \leq 7n + 3 < (k + 1)^3
   \]
   This simplifies to:
   \[
   k^3 \leq 7n + 2 < (k + 1)^3 \quad \text{and} \quad k^3 \leq 7n + 3 < (k + 1)^3
   \]
   Since \( 7n + 2 \) and \( 7n + 3 \) are consecutive integers, the only way for the floor values to differ is if \( 7n + 3 \) is exactly at the boundary of \( (k + 1)^3 \).

4. **Checking for Perfect Cubes**:
   The critical case is when \( 7n + 3 = (k + 1)^3 \). If \( 7n + 3 \) is a perfect cube, then:
   \[
   7n + 3 = m^3 \quad \text{for some integer } m
   \]
   Solving for \( n \), we get:
   \[
   7n = m^3 - 3 \quad \Rightarrow \quad n = \frac{m^3 - 3}{7}
   \]
   For \( n \) to be an integer, \( m^3 - 3 \) must be divisible by 7. 

5. **Modulo Analysis**:
   We need to check when \( m^3 \equiv 3 \pmod{7} \). The possible values of \( m \) modulo 7 are 0 through 6. Calculating \( m^3 \mod 7 \):
   \[
   \begin{aligned}
   0^3 &\equiv 0 \pmod{7}, \\
   1^3 &\equiv 1 \pmod{7}, \\
   2^3 &\equiv 8 \equiv 1 \pmod{7}, \\
   3^3 &\equiv 27 \equiv 6 \pmod{7}, \\
   4^3 &\equiv 64 \equiv 1 \pmod{7}, \\
   5^3 &\equiv 125 \equiv 6 \pmod{7}, \\
   6^3 &\equiv 216 \equiv 6 \pmod{7}.
   \end{aligned}
   \]
   None of these values are congruent to 3 modulo 7. Therefore, \( m^3 \equiv 3 \pmod{7} \) has no solutions.

6. **Conclusion**:
   Since \( 7n + 3 \) can never be a perfect cube, the equation \( \left\lfloor \sqrt[3]{7n + 2} \right\rfloor = \left\lfloor \sqrt[3]{7n + 3} \right\rfloor \) always holds.

The final answer is \(\boxed{0}\)