For every n = 2; 3; : : : , we put
$$A_n = \left(1 - \frac{1}{1+2}\right) X \left(1 - \frac{1}{1+2+3}\right)X \left(1 - \frac{1}{1+2+3+...+n}\right) $$
Determine all positive integer $ n (n \geq 2)$ such that $\frac{1}{A_n}$ is an integer.
1. We start with the given expression for \( A_n \):
   \[
   A_n = \left(1 - \frac{1}{1+2}\right) \times \left(1 - \frac{1}{1+2+3}\right) \times \cdots \times \left(1 - \frac{1}{1+2+3+\cdots+n}\right)
   \]
   
2. We need to simplify each term in the product. The \( k \)-th term in the product is:
   \[
   1 - \frac{1}{1+2+\cdots+k}
   \]
   The sum of the first \( k \) natural numbers is given by:
   \[
   1 + 2 + \cdots + k = \frac{k(k+1)}{2}
   \]
   Therefore, the \( k \)-th term becomes:
   \[
   1 - \frac{1}{\frac{k(k+1)}{2}} = 1 - \frac{2}{k(k+1)}
   \]

3. We can rewrite the term as:
   \[
   1 - \frac{2}{k(k+1)} = \frac{k(k+1) - 2}{k(k+1)} = \frac{k^2 + k - 2}{k(k+1)}
   \]
   Factoring the numerator, we get:
   \[
   \frac{k^2 + k - 2}{k(k+1)} = \frac{(k-1)(k+2)}{k(k+1)}
   \]

4. Thus, the product \( A_n \) becomes:
   \[
   A_n = \prod_{k=2}^n \frac{(k-1)(k+2)}{k(k+1)}
   \]

5. We can simplify this product by observing the telescoping nature of the product:
   \[
   A_n = \frac{1 \cdot 4}{2 \cdot 3} \times \frac{2 \cdot 5}{3 \cdot 4} \times \frac{3 \cdot 6}{4 \cdot 5} \times \cdots \times \frac{(n-1)(n+2)}{n(n+1)}
   \]
   Most terms in the numerator and denominator cancel out, leaving:
   \[
   A_n = \frac{n+2}{3n}
   \]

6. We need to determine for which \( n \geq 2 \), \( \frac{1}{A_n} \) is an integer. This means:
   \[
   \frac{1}{A_n} = \frac{3n}{n+2}
   \]
   must be an integer. Therefore, \( n+2 \) must divide \( 3n \).

7. We solve the divisibility condition:
   \[
   n+2 \mid 3n
   \]
   This implies there exists an integer \( k \) such that:
   \[
   3n = k(n+2)
   \]
   Rearranging, we get:
   \[
   3n = kn + 2k \implies 3n - kn = 2k \implies n(3-k) = 2k \implies n = \frac{2k}{3-k}
   \]

8. For \( n \) to be an integer, \( \frac{2k}{3-k} \) must be an integer. We test integer values of \( k \):
   - If \( k = 1 \):
     \[
     n = \frac{2 \cdot 1}{3-1} = \frac{2}{2} = 1 \quad (\text{not valid since } n \geq 2)
     \]
   - If \( k = 2 \):
     \[
     n = \frac{2 \cdot 2}{3-2} = \frac{4}{1} = 4 \quad (\text{valid})
     \]
   - If \( k = 3 \):
     \[
     n = \frac{2 \cdot 3}{3-3} = \frac{6}{0} \quad (\text{undefined})
     \]

9. Therefore, the only valid solution is \( n = 4 \).

The final answer is \( \boxed{n=4} \)