For each positive integer $n$, determine the smallest possible value of the polynomial
$$
W_n(x)=x^{2n}+2x^{2n-1}+3x^{2n-2}+\ldots + (2n-1)x^2+2nx.
$$
1. **Express the polynomial \( W_n(x) \) in a different form:**

   Given the polynomial:
   \[
   W_n(x) = x^{2n} + 2x^{2n-1} + 3x^{2n-2} + \ldots + (2n-1)x^2 + 2nx
   \]
   We can use TuZo's hint to rewrite it as:
   \[
   W_n(x) = \frac{x^{2n+2} - x^2}{(x-1)^2} - 2n \frac{x}{x-1}
   \]

2. **Analyze the polynomial for \( x \ge 0 \):**

   For \( x \ge 0 \), we need to determine the behavior of \( W_n(x) \). Notice that:
   \[
   W_n(x) \ge 0
   \]
   This is because each term in the polynomial is non-negative when \( x \ge 0 \).

3. **Evaluate the polynomial at \( x = -1 \):**

   For \( x = -1 \):
   \[
   W_n(-1) = (-1)^{2n} + 2(-1)^{2n-1} + 3(-1)^{2n-2} + \ldots + (2n-1)(-1)^2 + 2n(-1)
   \]
   Simplifying, we get:
   \[
   W_n(-1) = 1 - 2 + 3 - 4 + \ldots + (2n-1) - 2n
   \]
   This is an alternating series, and it sums to:
   \[
   W_n(-1) = -n
   \]
   Hence, the minimum value is attained for some \( x \le 0 \).

4. **Reparametrize and consider \( V_n(x) = W_n(-x) \):**

   Define:
   \[
   V_n(x) = W_n(-x) = \frac{x^{2n+2} - x^2}{(x+1)^2} - 2n \frac{x}{x+1}
   \]
   We need to analyze \( V_n(x) \) for \( x \ge 0 \).

5. **Apply the Bernoulli inequality:**

   For \( x \ge 0 \), the Bernoulli inequality states:
   \[
   x^{2n+2} \ge 1 + (n+1)(x^2 - 1)
   \]
   Using this inequality, we get:
   \[
   V_n(x) \ge \frac{1 + (n+1)(x^2 - 1) - x^2}{(x+1)^2} - 2n \frac{x}{x+1}
   \]
   Simplifying the right-hand side, we obtain:
   \[
   V_n(x) \ge -n
   \]

6. **Evaluate \( V_n(x) \) at \( x = 1 \):**

   For \( x = 1 \):
   \[
   V_n(1) = \frac{1^{2n+2} - 1^2}{(1+1)^2} - 2n \frac{1}{1+1} = \frac{1 - 1}{4} - n = -n
   \]
   Thus, the minimum value of \( V_n(x) \) is \( -n \).

7. **Conclusion:**

   Since \( V_n(x) \ge -n \) and \( V_n(1) = -n \), the minimum value of \( W_n(x) \) is \( -n \).

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