Let be two distinct natural numbers $ k_1 $ and $ k_2 $ and a sequence $ \left( x_n \right)_{n\ge 0} $ which satisfies
$$ x_nx_m +k_1k_2\le k_1x_n +k_2x_m,\quad\forall m,n\in\{ 0\}\cup\mathbb{N}. $$

Calculate $ \lim_{n\to\infty}\frac{n!\cdot (-1)^{1+n}\cdot x_n^2}{n^n} . $
1. **Given Inequality Analysis**:
   We start with the given inequality for the sequence \( (x_n)_{n \ge 0} \):
   \[
   x_n x_m + k_1 k_2 \le k_1 x_n + k_2 x_m, \quad \forall m, n \in \{0\} \cup \mathbb{N}.
   \]
   Let's consider the case when \( m = n \):
   \[
   x_n^2 + k_1 k_2 \le k_1 x_n + k_2 x_n.
   \]
   Simplifying this, we get:
   \[
   x_n^2 + k_1 k_2 \le (k_1 + k_2) x_n.
   \]
   Rearranging terms, we obtain:
   \[
   x_n^2 - (k_1 + k_2) x_n + k_1 k_2 \le 0.
   \]
   This can be rewritten as:
   \[
   (x_n - k_1)(x_n - k_2) \le 0.
   \]
   This inequality implies that \( x_n \) must lie between \( k_1 \) and \( k_2 \) (inclusive), since \( k_1 \) and \( k_2 \) are distinct natural numbers. Therefore, \( x_n \) is bounded:
   \[
   k_1 \le x_n \le k_2.
   \]

2. **Limit Calculation**:
   We need to calculate the limit:
   \[
   \lim_{n \to \infty} \frac{n! \cdot (-1)^{1+n} \cdot x_n^2}{n^n}.
   \]
   Since \( x_n \) is bounded, there exists a constant \( M \) such that \( |x_n| \le M \) for all \( n \). Therefore, \( x_n^2 \le M^2 \).

   Consider the expression:
   \[
   \left| \frac{n! \cdot (-1)^{1+n} \cdot x_n^2}{n^n} \right| = \frac{n! \cdot x_n^2}{n^n}.
   \]
   Using the bound \( x_n^2 \le M^2 \), we get:
   \[
   \left| \frac{n! \cdot x_n^2}{n^n} \right| \le \frac{n! \cdot M^2}{n^n}.
   \]
   We know from Stirling's approximation that \( n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \). Thus,
   \[
   \frac{n!}{n^n} \approx \frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right)^n}{n^n} = \sqrt{2 \pi n} \left( \frac{1}{e} \right)^n.
   \]
   As \( n \to \infty \), \( \left( \frac{1}{e} \right)^n \) approaches 0 exponentially fast. Therefore,
   \[
   \lim_{n \to \infty} \frac{n!}{n^n} = 0.
   \]
   Consequently,
   \[
   \lim_{n \to \infty} \frac{n! \cdot M^2}{n^n} = 0.
   \]
   Since \( \left| \frac{n! \cdot x_n^2}{n^n} \right| \le \frac{n! \cdot M^2}{n^n} \) and the right-hand side tends to 0, by the squeeze theorem, we have:
   \[
   \lim_{n \to \infty} \frac{n! \cdot x_n^2}{n^n} = 0.
   \]

The final answer is \(\boxed{0}\).