A Mediterranean polynomial has only real roots and it is of the form
\[ P(x) =  x^{10}-20x^9+135x^8+a_7x^7+a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0 \] with real coefficients $a_0\ldots,a_7$.  Determine the largest real number that occurs as a root of some Mediterranean polynomial.
[i](Proposed by Gerhard Woeginger, Austria)[/i]
1. **Define the polynomial and roots:**
   Consider a Mediterranean polynomial of the form:
   \[
   P(x) = x^{10} - 20x^9 + 135x^8 + a_7x^7 + a_6x^6 + a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0
   \]
   with real coefficients \(a_0, \ldots, a_7\). Let \(\alpha\) be one of its real roots, and let the other roots be \(x_1, x_2, \ldots, x_9\).

2. **Apply Vieta's formulas:**
   By Vieta's formulas, the sum of the roots of the polynomial is equal to the coefficient of \(x^9\) (with a sign change), and the sum of the products of the roots taken two at a time is equal to the coefficient of \(x^8\) (with a sign change). Therefore:
   \[
   \alpha + \sum_{i=1}^9 x_i = 20
   \]
   Let \(s = \sum_{i=1}^9 x_i\). Then:
   \[
   s = 20 - \alpha
   \]

3. **Sum of the products of the roots taken two at a time:**
   Let \(u = \sum_{1 \le i < j \le 9} x_i x_j\). By Vieta's formulas:
   \[
   u = 135 - s\alpha
   \]
   Substituting \(s = 20 - \alpha\):
   \[
   u = 135 - (20 - \alpha)\alpha = 135 - 20\alpha + \alpha^2
   \]

4. **Sum of the squares of the roots:**
   Let \(t = \sum_{i=1}^9 x_i^2\). Using the identity for the sum of squares:
   \[
   t = s^2 - 2u
   \]
   Substituting \(s = 20 - \alpha\) and \(u = 135 - 20\alpha + \alpha^2\):
   \[
   t = (20 - \alpha)^2 - 2(135 - 20\alpha + \alpha^2)
   \]
   Simplifying:
   \[
   t = 400 - 40\alpha + \alpha^2 - 270 + 40\alpha - 2\alpha^2 = 130 - \alpha^2
   \]

5. **Non-negativity of the sum of squared differences:**
   Consider the sum of the squared differences of the roots:
   \[
   \sum_{1 \le i < j \le 9} (x_i - x_j)^2
   \]
   This can be expressed as:
   \[
   \sum_{1 \le i < j \le 9} (x_i - x_j)^2 = 8t - 2u
   \]
   Substituting \(t = 130 - \alpha^2\) and \(u = 135 - 20\alpha + \alpha^2\):
   \[
   8t - 2u = 8(130 - \alpha^2) - 2(135 - 20\alpha + \alpha^2)
   \]
   Simplifying:
   \[
   8(130 - \alpha^2) - 2(135 - 20\alpha + \alpha^2) = 1040 - 8\alpha^2 - 270 + 40\alpha - 2\alpha^2 = 770 - 10\alpha^2 + 40\alpha
   \]

6. **Non-negativity condition:**
   Since the sum of squared differences must be non-negative:
   \[
   770 - 10\alpha^2 + 40\alpha \ge 0
   \]
   Dividing by 10:
   \[
   77 - \alpha^2 + 4\alpha \ge 0
   \]
   Rearranging:
   \[
   \alpha^2 - 4\alpha - 77 \le 0
   \]
   Solving the quadratic inequality:
   \[
   \alpha = \frac{4 \pm \sqrt{16 + 308}}{2} = \frac{4 \pm \sqrt{324}}{2} = \frac{4 \pm 18}{2}
   \]
   Thus:
   \[
   \alpha = 11 \quad \text{or} \quad \alpha = -7
   \]
   Since \(\alpha\) must be a real root, we consider the positive value:
   \[
   \alpha \le 11
   \]

7. **Verification:**
   If \(\alpha = 11\), then \(s = 20 - 11 = 9\) and \(x_1 = x_2 = \cdots = x_9 = 1\). This configuration satisfies the polynomial's conditions.

The final answer is \(\boxed{11}\).