Find all positive integers $n$ such that $n+ \left[ \frac{n}{6} \right] \ne \left[ \frac{n}{2} \right] + \left[ \frac{2n}{3} \right]$
To solve the problem, we need to find all positive integers \( n \) such that:
\[ n + \left\lfloor \frac{n}{6} \right\rfloor \ne \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{2n}{3} \right\rfloor \]

1. Define the function:
\[ f(n) = n + \left\lfloor \frac{n}{6} \right\rfloor - \left\lfloor \frac{n}{2} \right\rfloor - \left\lfloor \frac{2n}{3} \right\rfloor \]

2. We need to find when \( f(n) \ne 0 \).

3. Consider the periodicity of the floor functions involved. Notice that:
\[ f(n+6) = (n+6) + \left\lfloor \frac{n+6}{6} \right\rfloor - \left\lfloor \frac{n+6}{2} \right\rfloor - \left\lfloor \frac{2(n+6)}{3} \right\rfloor \]

4. Simplify \( f(n+6) \):
\[
f(n+6) = (n+6) + \left\lfloor \frac{n+6}{6} \right\rfloor - \left\lfloor \frac{n+6}{2} \right\rfloor - \left\lfloor \frac{2n+12}{3} \right\rfloor
\]
Since \(\left\lfloor \frac{n+6}{6} \right\rfloor = \left\lfloor \frac{n}{6} + 1 \right\rfloor = \left\lfloor \frac{n}{6} \right\rfloor + 1\), \(\left\lfloor \frac{n+6}{2} \right\rfloor = \left\lfloor \frac{n}{2} + 3 \right\rfloor = \left\lfloor \frac{n}{2} \right\rfloor + 3\), and \(\left\lfloor \frac{2n+12}{3} \right\rfloor = \left\lfloor \frac{2n}{3} + 4 \right\rfloor = \left\lfloor \frac{2n}{3} \right\rfloor + 4\), we get:
\[
f(n+6) = n + 6 + \left\lfloor \frac{n}{6} \right\rfloor + 1 - \left\lfloor \frac{n}{2} \right\rfloor - 3 - \left\lfloor \frac{2n}{3} \right\rfloor - 4
\]
\[
f(n+6) = n + \left\lfloor \frac{n}{6} \right\rfloor - \left\lfloor \frac{n}{2} \right\rfloor - \left\lfloor \frac{2n}{3} \right\rfloor + 7 - 7
\]
\[
f(n+6) = f(n)
\]

5. Since \( f(n) \) is periodic with period 6, we only need to check \( n \) modulo 6. We will check \( n = 0, 1, 2, 3, 4, 5 \) modulo 6.

6. Calculate \( f(n) \) for \( n = 0, 1, 2, 3, 4, 5 \):
   - For \( n = 0 \):
     \[
     f(0) = 0 + \left\lfloor \frac{0}{6} \right\rfloor - \left\lfloor \frac{0}{2} \right\rfloor - \left\lfloor \frac{2 \cdot 0}{3} \right\rfloor = 0
     \]
   - For \( n = 1 \):
     \[
     f(1) = 1 + \left\lfloor \frac{1}{6} \right\rfloor - \left\lfloor \frac{1}{2} \right\rfloor - \left\lfloor \frac{2 \cdot 1}{3} \right\rfloor = 1 + 0 - 0 - 0 = 1
     \]
   - For \( n = 2 \):
     \[
     f(2) = 2 + \left\lfloor \frac{2}{6} \right\rfloor - \left\lfloor \frac{2}{2} \right\rfloor - \left\lfloor \frac{2 \cdot 2}{3} \right\rfloor = 2 + 0 - 1 - 1 = 0
     \]
   - For \( n = 3 \):
     \[
     f(3) = 3 + \left\lfloor \frac{3}{6} \right\rfloor - \left\lfloor \frac{3}{2} \right\rfloor - \left\lfloor \frac{2 \cdot 3}{3} \right\rfloor = 3 + 0 - 1 - 2 = 0
     \]
   - For \( n = 4 \):
     \[
     f(4) = 4 + \left\lfloor \frac{4}{6} \right\rfloor - \left\lfloor \frac{4}{2} \right\rfloor - \left\lfloor \frac{2 \cdot 4}{3} \right\rfloor = 4 + 0 - 2 - 2 = 0
     \]
   - For \( n = 5 \):
     \[
     f(5) = 5 + \left\lfloor \frac{5}{6} \right\rfloor - \left\lfloor \frac{5}{2} \right\rfloor - \left\lfloor \frac{2 \cdot 5}{3} \right\rfloor = 5 + 0 - 2 - 3 = 0
     \]

7. From the calculations, we see that \( f(n) \ne 0 \) only when \( n \equiv 1 \pmod{6} \).

Conclusion:
The positive integers \( n \) that satisfy the given condition are of the form \( 6k + 1 \) for \( k \in \mathbb{Z}_{\geq 0} \).

The final answer is \( \boxed{ n = 6k + 1 } \) for \( k \in \mathbb{Z}_{\geq 0} \).