For a given positive integer $n(\ge 2)$, find maximum positive integer $A$ such that a polynomial with integer coefficients $P(x)$ with degree $n$ that satisfies the condition below exists.


$\bullet P(1), P(2), \cdots P(A)$ is a multiple of $A$

$\bullet P(0) = 0$ and the coefficient of the first term of $P(x)$ is $1$, In other words, $P(x)$ has the following form:
$$P(x) = c_nx^n + c_{n-1}x^{n-1}+\dots+c_2x^2+x \ \ (c_n \neq 0)$$
1. **Claim**: The maximum positive integer \( A \) such that a polynomial \( P(x) \) with integer coefficients and degree \( n \) satisfies the given conditions is \( n! \).

2. **Constructing a Polynomial**: Consider the polynomial \( P(x) = x(x+1)(x+2)\cdots(x+n-1) \). This polynomial has the following properties:
   - It is a monic polynomial of degree \( n \).
   - The coefficient of the linear term is 1.
   - \( P(0) = 0 \).

3. **Verifying the Conditions**:
   - For \( P(i) \) where \( i = 1, 2, \ldots, n \):
     \[
     P(i) = i(i+1)(i+2)\cdots(i+n-1)
     \]
     This can be rewritten using the factorial notation:
     \[
     P(i) = \frac{(i+n-1)!}{(i-1)!}
     \]
     Since \( (i+n-1)! \) is divisible by \( n! \), it follows that \( P(i) \) is a multiple of \( n! \) for \( i = 1, 2, \ldots, n \).

4. **Upper Bound for \( A \)**:
   - We need to show that \( A \leq n! \). Consider the \( n \)-th finite difference of \( P(x) \), denoted by \( \Delta^n P(x) \).
   - For a polynomial \( P(x) \) of degree \( n \), the \( n \)-th finite difference \( \Delta^n P(x) \) is a constant and equals \( n! \).
   - By the given condition, \( P(z) \) is a multiple of \( A \) for all positive integers \( z \). Therefore, \( A \) must divide \( \Delta^n P(0) \), which is \( n! \).

5. **Conclusion**: Since \( A \) must divide \( n! \) and we have constructed a polynomial where \( P(i) \) is a multiple of \( n! \) for \( i = 1, 2, \ldots, n \), the maximum value of \( A \) is \( n! \).

\(\blacksquare\)

The final answer is \( \boxed{ n! } \)