Find all polynomials $P_n(x)$ of the form
$$P_n(x)=n!x^n+a_{n-1}x^{n-1}+\ldots+a_1x+(-1)^nn(n+1),$$with integer coefficients, such that its roots $x_1,x_2,\ldots,x_n$ satisfy $k\le x_k\le k+1$ for $k=1,2,\ldots,n$.
1. **Applying Vieta's Theorem:**
   According to Vieta's formulas, the product of the roots of the polynomial \( P_n(x) \) is given by:
   \[
   \prod_{i=1}^n x_i = \frac{(-1)^n \cdot \text{constant term}}{n!}
   \]
   For the polynomial \( P_n(x) = n!x^n + a_{n-1}x^{n-1} + \ldots + a_1x + (-1)^n n(n+1) \), the constant term is \( (-1)^n n(n+1) \). Therefore:
   \[
   \prod_{i=1}^n x_i = \frac{(-1)^n \cdot (-1)^n n(n+1)}{n!} = \frac{n(n+1)}{n!}
   \]

2. **Analyzing the Product of Roots:**
   Given the condition \( k \le x_k \le k+1 \) for \( k = 1, 2, \ldots, n \), we know that each root \( x_k \) is at least \( k \). Therefore:
   \[
   \prod_{i=1}^n x_i \ge \prod_{i=1}^n i = n!
   \]
   Combining this with the earlier expression for the product of the roots, we get:
   \[
   \frac{n(n+1)}{n!} \ge n!
   \]
   Simplifying this inequality:
   \[
   n(n+1) \ge (n!)^2
   \]

3. **Checking the Inequality for Small Values of \( n \):**
   - For \( n = 1 \):
     \[
     1(1+1) = 2 \quad \text{and} \quad (1!)^2 = 1
     \]
     Clearly, \( 2 \ge 1 \) holds.
   - For \( n = 2 \):
     \[
     2(2+1) = 6 \quad \text{and} \quad (2!)^2 = 4
     \]
     Clearly, \( 6 \ge 4 \) holds.
   - For \( n = 3 \):
     \[
     3(3+1) = 12 \quad \text{and} \quad (3!)^2 = 36
     \]
     Clearly, \( 12 \not\ge 36 \) does not hold.

   Therefore, the inequality \( n(n+1) \ge (n!)^2 \) holds only for \( n = 1 \) and \( n = 2 \).

4. **Finding the Polynomials for \( n = 1 \) and \( n = 2 \):**
   - For \( n = 1 \):
     \[
     P_1(x) = 1!x^1 + (-1)^1 \cdot 1(1+1) = x - 2
     \]
     The root is \( x = 2 \), which satisfies \( 1 \le x \le 2 \).

   - For \( n = 2 \):
     \[
     P_2(x) = 2!x^2 + a_1x + (-1)^2 \cdot 2(2+1) = 2x^2 + a_1x + 6
     \]
     The roots \( x_1 \) and \( x_2 \) must satisfy \( 1 \le x_1 \le 2 \) and \( 2 \le x_2 \le 3 \). Using Vieta's formulas:
     \[
     x_1 + x_2 = -\frac{a_1}{2} \quad \text{and} \quad x_1 x_2 = \frac{6}{2} = 3
     \]
     Since \( x_1 + x_2 \in [3, 5] \), we have:
     \[
     -\frac{a_1}{2} \in [3, 5] \implies a_1 \in [-10, -6] \cap \mathbb{Z}
     \]
     Therefore, \( a_1 \) can be any integer in the set \( \{-10, -9, -8, -7, -6\} \).

The final answer is \( \boxed{ P_1(x) = x - 2 } \) and \( P_2(x) = 2x^2 + a_1x + 6 \) where \( a_1 \in \{-10, -9, -8, -7, -6\} \).