Let $p$ be a fixed odd prime. Find the minimum positive value of $E_{p}(x,y) = \sqrt{2p}-\sqrt{x}-\sqrt{y}$ where $x,y \in \mathbb{Z}_{+}$.
1. We start with the given function \( E_p(x, y) = \sqrt{2p} - \sqrt{x} - \sqrt{y} \) and aim to find its minimum positive value for \( x, y \in \mathbb{Z}_+ \).

2. To find the minimum positive value of \( E_p(x, y) \), we need to maximize \( \sqrt{x} + \sqrt{y} \) under the constraint \( \sqrt{x} + \sqrt{y} < \sqrt{2p} \).

3. Consider the inequality \( \sqrt{x} + \sqrt{y} \leq \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} \). We need to prove this inequality.

4. Let \( a = \sqrt{x} \) and \( b = \sqrt{y} \). Then the inequality becomes:
   \[
   a + b \leq \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}}
   \]

5. Squaring both sides, we get:
   \[
   (a + b)^2 \leq \left( \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} \right)^2
   \]

6. Expanding both sides, we have:
   \[
   a^2 + 2ab + b^2 \leq \left( \sqrt{\frac{p-1}{2}} \right)^2 + 2 \sqrt{\frac{p-1}{2}} \sqrt{\frac{p+1}{2}} + \left( \sqrt{\frac{p+1}{2}} \right)^2
   \]

7. Simplifying the right-hand side:
   \[
   a^2 + 2ab + b^2 \leq \frac{p-1}{2} + \frac{p+1}{2} + 2 \sqrt{\frac{p-1}{2}} \sqrt{\frac{p+1}{2}}
   \]

8. Notice that:
   \[
   \frac{p-1}{2} + \frac{p+1}{2} = \frac{p-1 + p+1}{2} = \frac{2p}{2} = p
   \]

9. Therefore, the inequality becomes:
   \[
   a^2 + 2ab + b^2 \leq p + 2 \sqrt{\frac{p-1}{2}} \sqrt{\frac{p+1}{2}}
   \]

10. We need to show that \( 2 \sqrt{\frac{p-1}{2}} \sqrt{\frac{p+1}{2}} \leq 2ab \). This is true if:
    \[
    \sqrt{\frac{p-1}{2}} \sqrt{\frac{p+1}{2}} \leq ab
    \]

11. Since \( a = \sqrt{x} \) and \( b = \sqrt{y} \), we have:
    \[
    ab = \sqrt{x} \sqrt{y} = \sqrt{xy}
    \]

12. Therefore, the inequality becomes:
    \[
    \sqrt{\frac{p-1}{2}} \sqrt{\frac{p+1}{2}} \leq \sqrt{xy}
    \]

13. Squaring both sides, we get:
    \[
    \frac{(p-1)(p+1)}{4} \leq xy
    \]

14. Simplifying the left-hand side:
    \[
    \frac{p^2 - 1}{4} \leq xy
    \]

15. Therefore, we have:
    \[
    xy \geq \frac{p^2 - 1}{4}
    \]

16. Since \( x \) and \( y \) are positive integers, the minimum value of \( xy \) that satisfies this inequality is \( \left\lceil \frac{p^2 - 1}{4} \right\rceil \).

17. Thus, the minimum positive value of \( E_p(x, y) \) is achieved when \( \sqrt{x} + \sqrt{y} \) is as close as possible to \( \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} \).

The final answer is \( \boxed{ \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} } \).