Let $ n>0$ be a natural number. Determine all the polynomials of degree $ 2n$ with real coefficients in the form 
$ P(X)\equal{}X^{2n}\plus{}(2n\minus{}10)X^{2n\minus{}1}\plus{}a_2X^{2n\minus{}2}\plus{}...\plus{}a_{2n\minus{}2}X^2\plus{}(2n\minus{}10)X\plus{}1$,
if it is known that all the roots of them are positive reals.

[i]Proposer[/i]: [b]Baltag Valeriu[/b]
1. **Given Polynomial and Roots:**
   We are given a polynomial of degree \(2n\) with real coefficients:
   \[
   P(X) = X^{2n} + (2n-10)X^{2n-1} + a_2X^{2n-2} + \cdots + a_{2n-2}X^2 + (2n-10)X + 1
   \]
   and it is known that all the roots of this polynomial are positive reals. Let the roots be \(b_1, b_2, \ldots, b_{2n}\).

2. **Product of Roots:**
   By Vieta's formulas, the product of the roots of the polynomial is given by:
   \[
   b_1 b_2 \cdots b_{2n} = 1
   \]
   This implies that none of the roots \(b_i\) can be zero.

3. **Sum of Roots:**
   The sum of the roots, again by Vieta's formulas, is:
   \[
   b_1 + b_2 + \cdots + b_{2n} = 10 - 2n
   \]

4. **Sum of Reciprocals of Roots:**
   Since the product of the roots is 1, the sum of the reciprocals of the roots is:
   \[
   \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_{2n}} = 10 - 2n
   \]

5. **Applying AM-GM Inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \frac{b_1 + b_2 + \cdots + b_{2n}}{2n} \geq \sqrt[2n]{b_1 b_2 \cdots b_{2n}} = 1
   \]
   This simplifies to:
   \[
   \frac{10 - 2n}{2n} \geq 1 \implies 10 - 2n \geq 2n \implies 10 \geq 4n \implies n \leq 2.5
   \]
   Since \(n\) is a natural number, the possible values for \(n\) are 1 and 2.

6. **Case \(n = 1\):**
   For \(n = 1\), the polynomial becomes:
   \[
   P(X) = X^2 - 8X + 1
   \]
   The roots of this polynomial are:
   \[
   X = \frac{8 \pm \sqrt{64 - 4}}{2} = 4 \pm \sqrt{15}
   \]
   Both roots are positive, so this polynomial satisfies the given conditions.

7. **Case \(n = 2\):**
   For \(n = 2\), the polynomial becomes:
   \[
   P(X) = X^4 - 6X^3 + a_2X^2 - 6X + 1
   \]
   We need to find the possible values of \(a_2\) such that all roots are positive.

8. **Roots and Symmetry:**
   Let the roots be \(b_1, b_2, b_3, b_4\). Since the product of the roots is 1, we can assume without loss of generality that \(b_1 = \frac{1}{b_2}\) and \(b_3 = \frac{1}{b_4}\).

9. **Sum of Roots:**
   The sum of the roots is:
   \[
   b_1 + \frac{1}{b_1} + b_3 + \frac{1}{b_3} = 6
   \]
   Let \(b_1 + \frac{1}{b_1} = m\) and \(b_3 + \frac{1}{b_3} = k\). Then:
   \[
   m + k = 6
   \]

10. **Finding \(a_2\):**
    We need to find \(a_2\) in terms of \(m\) and \(k\):
    \[
    a_2 = b_1 b_2 + b_1 b_3 + b_1 b_4 + b_2 b_3 + b_2 b_4 + b_3 b_4
    \]
    Since \(b_1 b_2 = 1\), we have:
    \[
    a_2 = 1 + b_1 b_3 + \frac{b_3}{b_1} + \frac{b_1}{b_3} + \frac{1}{b_1 b_3} + 1
    \]
    \[
    a_2 = 2 + (b_1 + \frac{1}{b_1})(b_3 + \frac{1}{b_3}) = 2 + mk
    \]
    Since \(m + k = 6\), we have:
    \[
    m = 3 \pm \sqrt{11 - a_2}
    \]
    The restrictions on \(a_2\) are:
    \[
    3 \pm \sqrt{11 - a_2} \geq 2 \implies \sqrt{11 - a_2} \leq 1 \implies 11 - a_2 \leq 1 \implies a_2 \geq 10
    \]
    Also, \(11 - a_2 \geq 0 \implies a_2 \leq 11\).

11. **Conclusion for \(n = 2\):**
    Therefore, for \(n = 2\), the polynomial is:
    \[
    P(X) = X^4 - 6X^3 + a_2X^2 - 6X + 1
    \]
    with \(10 \leq a_2 \leq 11\).

The final answer is \( \boxed{ P(X) = X^2 - 8X + 1 } \) for \(n = 1\) and \(P(X) = X^4 - 6X^3 + a_2X^2 - 6X + 1\) with \(10 \leq a_2 \leq 11\) for \(n = 2\).