Determine all polynomials $P(x)$ with integer coefficients which satisfies $P(n)\mid n!+2$ for all postive integer $n$.
1. **Claim**: The only polynomials \( P(x) \) with integer coefficients that satisfy \( P(n) \mid n! + 2 \) for all positive integers \( n \) are \( P(x) = \pm 1 \).

2. **Verification for \( P(x) = \pm 1 \)**:
   - If \( P(x) = 1 \), then \( P(n) = 1 \) for all \( n \), and clearly \( 1 \mid n! + 2 \) for all \( n \).
   - If \( P(x) = -1 \), then \( P(n) = -1 \) for all \( n \), and clearly \( -1 \mid n! + 2 \) for all \( n \).

3. **Assume \( P(x) \) is nonconstant**:
   - Since \( P(x) \) is a polynomial with integer coefficients, there exists an integer \( k > 3 \) such that \( |P(k)| > 2 \).
   - Let \( p \) be any prime dividing \( P(k) \). Then \( p \mid P(k) \).

4. **Using the property of polynomials**:
   - By the properties of polynomials, \( P(k+p) - P(k) \) is divisible by \( p \):
     \[
     (k+p) - k \mid P(k+p) - P(k) \implies p \mid P(k+p) - P(k) \implies p \mid P(k+p)
     \]
   - Therefore, \( p \mid P(k+p) \).

5. **Implication for factorials**:
   - Since \( p \mid P(k+p) \), it follows that \( p \mid (k+p)! + 2 \):
     \[
     p \mid P(k+p) \implies p \mid (k+p)! + 2
     \]
   - This implies \( p \mid 2 \), so \( p = 2 \).

6. **Form of \( P(k) \)**:
   - Since \( p = 2 \), \( P(k) \) must be a power of 2, say \( P(k) = 2^m \) for some natural number \( m \).

7. **Contradiction**:
   - For \( P(k) = 2^m \) to divide \( k! + 2 \), we must have \( 2^m \mid k! + 2 \).
   - However, for \( k \geq 4 \), \( k! \equiv 0 \pmod{4} \), so \( k! + 2 \equiv 2 \pmod{4} \).
   - This implies \( 2^m \mid 2 \), so \( m = 1 \), and \( P(k) = 2 \).

8. **Conclusion for nonconstant \( P(x) \)**:
   - If \( P(x) \) is nonconstant, then \( P(k) = 2 \) for infinitely many \( k \), which implies \( P(x) \equiv 2 \).
   - This is a contradiction because \( P(x) \) is assumed to be nonconstant.

9. **Conclusion for constant \( P(x) \)**:
   - If \( P(x) \) is constant, say \( P(x) \equiv c \), then \( c \mid 3 \) and \( c \mid 4 \), so \( c = \pm 1 \).

Thus, the only polynomials \( P(x) \) with integer coefficients that satisfy \( P(n) \mid n! + 2 \) for all positive integers \( n \) are \( P(x) = \pm 1 \).

The final answer is \( \boxed{ P(x) = \pm 1 } \).