Let $ n\geq 3 $ be an integer and let $ x_1,x_2,\ldots,x_{n-1} $ be nonnegative integers such that 

\begin{eqnarray*} \ x_1 + x_2 + \cdots + x_{n-1} &=& n \\ x_1 + 2x_2 + \cdots + (n-1)x_{n-1} &=& 2n-2. \end{eqnarray*}

Find the minimal value of $ F(x_1,x_2,\ldots,x_n) = \sum_{k=1}^{n-1} k(2n-k)x_k $.
**
   We need to find nonnegative integers \(x_1, x_2, \ldots, x_{n-1}\) that satisfy these equations. Let's try \(x_1 = n-1\) and \(x_{n-1} = 1\), and all other \(x_k = 0\) for \(k \neq 1, n-1\).

   Substituting these values:
   \[
   x_1 = n-1, \quad x_{n-1} = 1, \quad x_2 = x_3 = \cdots = x_{n-2} = 0
   \]

   Check the constraints:
   \[
   (n-1) + 0 + 0 + \cdots + 0 + 1 = n
   \]
   \[
   (n-1) \cdot 1 + 0 \cdot 2 + 0 \cdot 3 + \cdots + 0 \cdot (n-2) + 1 \cdot (n-1) = n-1 + n-1 = 2n-2
   \]

   Both constraints are satisfied.

5. **Calculating \(F\):**
   Now, we calculate \(F\) for these values:
   \[
   F = \sum_{k=1}^{n-1} k(2n-k)x_k
   \]
   Substituting \(x_1 = n-1\) and \(x_{n-1} = 1\):
   \[
   F = 1 \cdot (2n-1) \cdot (n-1) + (n-1) \cdot (2n-(n-1)) \cdot 1
   \]
   \[
   F = (2n-1)(n-1) + (n-1)(n+1)
   \]
   \[
   F = (2n-1)(n-1) + (n-1)(n+1)
   \]
   \[
   F = (n-1)(2n-1) + (n-1)(n+1)
   \]
   \[
   F = (n-1)[(2n-1) + (n+1)]
   \]
   \[
   F = (n-1)(3n)
   \]
   \[
   F = 3n(n-1)
   \]

The final answer is \(\boxed{3n(n-1)}\).