Let $p$ be an odd prime. Determine the positive integers $x$ and $y$ with $x\leq y$ for which the number $\sqrt{2p}-\sqrt{x}-\sqrt{y}$ is non-negative and as small as possible.
1. **Claim**: The required integers \( x \) and \( y \) are \( x = \frac{p-1}{2} \) and \( y = \frac{p+1}{2} \).

2. **Verification**: First, we verify that \( \left(\sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}}\right)^2 = p + \sqrt{p^2 - 1} \).

   \[
   \left(\sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}}\right)^2 = \frac{p-1}{2} + \frac{p+1}{2} + 2 \sqrt{\frac{p-1}{2}} \sqrt{\frac{p+1}{2}}
   \]
   \[
   = p + \sqrt{(p-1)(p+1)} = p + \sqrt{p^2 - 1}
   \]

3. **Non-negativity**: We need to show that \( \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}} \) is non-negative.

   \[
   \sqrt{2p} \geq \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} \implies 2p \geq \left(\sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}}\right)^2
   \]
   \[
   \implies 2p \geq p + \sqrt{p^2 - 1} \implies p \geq \sqrt{p^2 - 1} \implies p^2 \geq p^2 - 1
   \]

   This is true, so the non-negativity condition is satisfied.

4. **Minimality**: We need to show that \( \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}} \) is the smallest possible non-negative value for \( \sqrt{2p} - \sqrt{x} - \sqrt{y} \) with \( x \leq y \).

   Assume \( x \) and \( y \) are positive integers with \( x \leq y \) and:
   \[
   0 \leq \sqrt{2p} - \sqrt{x} - \sqrt{y} \leq \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}}
   \]

   This implies:
   \[
   \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} \leq \sqrt{x} + \sqrt{y} \leq \sqrt{2p}
   \]

5. **Squaring the inequality**:
   \[
   \left(\sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}}\right)^2 \leq \left(\sqrt{x} + \sqrt{y}\right)^2 \leq 2p
   \]
   \[
   \implies p + \sqrt{p^2 - 1} \leq x + y + 2\sqrt{xy} \leq 2p
   \]

6. **Using QM-AM inequality**:
   \[
   x + y = \left(\sqrt{x}\right)^2 + \left(\sqrt{y}\right)^2 \geq 2 \left(\frac{\sqrt{x} + \sqrt{y}}{2}\right)^2 = \frac{1}{2} \left(\sqrt{x} + \sqrt{y}\right)^2
   \]
   \[
   \geq \frac{1}{2} \left(\sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}}\right)^2 = \frac{1}{2} \left(p + \sqrt{p^2 - 1}\right)
   \]

   Since \( \sqrt{p^2 - 1} > \sqrt{p^2 - 2p + 1} = p - 1 \), this yields:
   \[
   x + y > \frac{1}{2} \left(p + (p - 1)\right) = p - \frac{1}{2}
   \]

   Since \( x \) and \( y \) are integers, this entails \( x + y \geq p \).

7. **Expressing \( x \) and \( y \)**:
   \[
   x + y = p + s \quad \text{for some integer } s \text{ satisfying } 0 \leq s \leq p - 1
   \]
   \[
   x = \frac{p + s}{2} + \frac{k}{2}, \quad y = \frac{p + s}{2} - \frac{k}{2}
   \]

8. **Simplifying the inequality**:
   \[
   p + \sqrt{p^2 - 1} \leq p + s + \sqrt{(p + s + k)(p + s - k)} \leq 2p
   \]
   \[
   \implies \sqrt{p^2 - 1} - s \leq \sqrt{(p + s)^2 - k^2} \leq p - s
   \]

   Since \( \sqrt{p^2 - 1} > \sqrt{p^2 - 2p + 1} = p - 1 \), the left-hand side is positive. Squaring the inequality:
   \[
   (\sqrt{p^2 - 1} - s)^2 \leq (p + s)^2 - k^2 \leq (p - s)^2
   \]
   \[
   \implies (p^2 - 1) + s^2 - 2\sqrt{p^2 - 1} \cdot s \leq p^2 + s^2 + 2ps - k^2 \leq p^2 + s^2 - 2ps
   \]
   \[
   \implies -1 - 2\sqrt{p^2 - 1} \cdot s \leq 2ps - k^2 \leq -2ps
   \]
   \[
   \implies 1 + 2\sqrt{p^2 - 1} \cdot s \geq k^2 - 2ps \geq 2ps
   \]
   \[
   \implies 1 + 2\sqrt{p^2 - 1} \cdot s + 2ps \geq k^2 \geq 4ps
   \]

   Since \( \sqrt{p^2 - 1} < \sqrt{p^2} = p \), we have:
   \[
   1 + 2p \cdot s + 2ps > k^2 \geq 4ps \implies 1 + 4ps > k^2 \geq 4ps
   \]

   Since \( k^2 \) is an integer, we have \( k^2 = 4ps \). Thus, \( p \mid k^2 \), and since \( p \) is prime, \( p \mid k \), so \( p^2 \mid k^2 \). Thus, \( p^2 \mid 4ps \), so \( p \mid 4s \). Since \( p \) is odd, this yields \( p \mid s \). Since \( 0 \leq s \leq p - 1 \), this leads to \( s = 0 \).

9. **Final values**:
   \[
   x = \frac{p + s}{2} + \frac{k}{2} = \frac{p}{2} + \frac{k}{2}, \quad y = \frac{p + s}{2} - \frac{k}{2} = \frac{p}{2} - \frac{k}{2}
   \]

   Since \( k \leq 0 \), this yields either \( k = 0 \) or \( k = -1 \). For \( k = 0 \), \( x = \frac{p}{2} \), which is impossible since \( p \) is odd. Thus, \( k = -1 \), so:
   \[
   x = \frac{p}{2} + \frac{k}{2} = \frac{p-1}{2}, \quad y = \frac{p}{2} - \frac{k}{2} = \frac{p+1}{2}
   \]

The final answer is \( \boxed{ x = \frac{p-1}{2} } \) and \( y = \frac{p+1}{2} \).