Let $n$ be a positive integer. Find all positive integers $m$ for which there exists a polynomial $f(x) = a_{0} + \cdots + a_{n}x^{n} \in \mathbb{Z}[X]$ ($a_{n} \not= 0$) such that $\gcd(a_{0},a_{1},\cdots,a_{n},m)=1$ and $m|f(k)$ for each $k \in \mathbb{Z}$.
1. **Understanding the Problem:**
   We need to find all positive integers \( m \) such that there exists a polynomial \( f(x) = a_0 + a_1 x + \cdots + a_n x^n \in \mathbb{Z}[X] \) with \( \gcd(a_0, a_1, \ldots, a_n, m) = 1 \) and \( m \) divides \( f(k) \) for all integers \( k \).

2. **Lattice of Polynomials:**
   The "lattice" of polynomials formed by \( \binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{n} \) are all polynomials that have degree at most \( n \) and are integers at integer points. This means that any polynomial \( f(x) \) that is integer-valued at integer points can be expressed as an integer linear combination of these binomial coefficients.

3. **Form of \( \frac{f(x)}{m} \):**
   Since \( m \) divides \( f(k) \) for all integers \( k \), we can write \( \frac{f(x)}{m} \) as a polynomial with integer coefficients. This implies that \( f(x) \) must be of the form \( m \cdot g(x) \) where \( g(x) \) is a polynomial with integer coefficients.

4. **Leading Coefficient Analysis:**
   The leading coefficient of \( f(x) \) is \( a_n \). For \( \frac{f(x)}{m} \) to be a polynomial with integer coefficients, \( m \) must divide \( a_n \). Since \( \gcd(a_0, a_1, \ldots, a_n, m) = 1 \), \( m \) must divide the factorial of the degree of the polynomial, \( n! \). This is because the leading coefficient of the polynomial \( x(x-1)(x-2)\ldots(x-n+1) \) is \( n! \).

5. **Inductive Proof:**
   - **Base Case (Degree 0):** Polynomials of the form \( k \cdot \binom{x}{0} \) are the only 0-degree integer-valued polynomials. Hence, for \( n = 0 \), \( m \) can be any positive integer.
   - **Inductive Step:** Assume the statement is true for polynomials of degree \( n-1 \). For a polynomial \( f(x) \) of degree \( n \), consider \( f(x) - f(x-1) \). This polynomial is of degree \( n-1 \) and must also be integer-valued at integer points. By the inductive hypothesis, \( m \) must divide \( (n-1)! \). Therefore, \( m \) must divide \( n! \).

6. **Conclusion:**
   The necessary and sufficient condition for \( m \) is that \( m \) must divide \( n! \).

The final answer is \( \boxed{ m } \) must divide \( n! \).