Let $p$ be an odd prime. Determine positive integers $x$ and $y$ for which $x \le y$ and $\sqrt{2p}-\sqrt{x}-\sqrt{y}$ is nonnegative and as small as possible.
1. **Claim**: The required integers \( x \) and \( y \) are \( x = \frac{p-1}{2} \) and \( y = \frac{p+1}{2} \). We need to show that the number \( \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}} \) is the smallest possible nonnegative value of the term \( \sqrt{2p} - \sqrt{x} - \sqrt{y} \) for positive integers \( x \) and \( y \) with \( x \leq y \).

2. **Assertion 1**: The numbers \( \frac{p-1}{2} \) and \( \frac{p+1}{2} \) are positive integers and satisfy \( \frac{p-1}{2} \leq \frac{p+1}{2} \). Also, the number \( \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}} \) is nonnegative.

   - Since \( p \) is an odd prime, \( p-1 \) and \( p+1 \) are even, making \( \frac{p-1}{2} \) and \( \frac{p+1}{2} \) integers.
   - Clearly, \( \frac{p-1}{2} \leq \frac{p+1}{2} \).
   - We need to show \( \sqrt{2p} \geq \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} \).

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

     \[
     \sqrt{2p} \geq \sqrt{\frac{p-1}{2}} + \sqrt{\frac{p+1}{2}} \Longleftrightarrow 2p \geq p + \sqrt{p^2 - 1} \Longleftrightarrow p \geq \sqrt{p^2 - 1} \Longleftrightarrow p^2 \geq p^2 - 1
     \]

     This is true, so Assertion 1 is proven.

3. **Assertion 2**: If \( x \) and \( y \) are positive integers satisfying \( x \leq y \) and 

   \[
   0 \leq \sqrt{2p} - \sqrt{x} - \sqrt{y} \leq \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}},
   \]

   then \( x = \frac{p-1}{2} \) and \( y = \frac{p+1}{2} \).

   - The inequality chain rewrites as:

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

   - Squaring this inequality:

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

     \[
     p + \sqrt{p^2 - 1} \leq x + y + 2 \sqrt{xy} \leq 2p
     \]

   - Using the QM-AM inequality for \( \sqrt{x} \) and \( \sqrt{y} \):

     \[
     x + y = (\sqrt{x})^2 + (\sqrt{y})^2 \geq 2 \left( \frac{\sqrt{x} + \sqrt{y}}{2} \right)^2 = \frac{1}{2} \left( \sqrt{x} + \sqrt{y} \right)^2
     \]

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

     Since \( \sqrt{p^2 - 1} > \sqrt{p^2 - 2p + 1} = p - 1 \), this yields:

     \[
     x + y > \frac{1}{2} (p + (p - 1)) = p - \frac{1}{2}
     \]

     Since \( x \) and \( y \) are integers, this entails \( x + y \geq p \). So we have \( p \leq x + y < 2p \). Thus, \( x + y = p + s \) for some integer \( s \) satisfying \( 0 \leq s \leq p - 1 \). Denoting \( x - y = k \), we note that \( k \leq 0 \) (since \( x \leq y \)) and:

     \[
     x = \frac{x + y}{2} + \frac{x - y}{2} = \frac{p + s}{2} + \frac{k}{2}
     \]

     \[
     y = \frac{x + y}{2} - \frac{x - y}{2} = \frac{p + s}{2} - \frac{k}{2}
     \]

   - The inequality \( p + \sqrt{p^2 - 1} \leq x + y + 2 \sqrt{xy} \leq 2p \) becomes:

     \[
     p + \sqrt{p^2 - 1} \leq p + s + 2 \sqrt{\frac{p + s}{2} + \frac{k}{2}} \cdot \sqrt{\frac{p + s}{2} - \frac{k}{2}} \leq 2p
     \]

     \[
     \Longleftrightarrow \sqrt{p^2 - 1} \leq s + \sqrt{(p + s + k)(p + s - k)} \leq p - s
     \]

     \[
     \Longleftrightarrow \sqrt{p^2 - 1} - s \leq \sqrt{(p + s)^2 - k^2} \leq p - s
     \]

   - Since \( \sqrt{p^2 - 1} > \sqrt{p^2 - 2p + 1} = p - 1 \geq s \), the left-hand side of this inequality, \( \sqrt{p^2 - 1} - s \), is positive; thus, we can square this inequality:

     \[
     (\sqrt{p^2 - 1} - s)^2 \leq (p + s)^2 - k^2 \leq (p - s)^2
     \]

     \[
     \Longleftrightarrow (p^2 - 1) + s^2 - 2 \sqrt{p^2 - 1} \cdot s \leq p^2 + s^2 + 2ps - k^2 \leq p^2 + s^2 - 2ps
     \]

     \[
     \Longleftrightarrow -1 - 2 \sqrt{p^2 - 1} \cdot s \leq 2ps - k^2 \leq -2ps
     \]

     \[
     \Longleftrightarrow 1 + 2 \sqrt{p^2 - 1} \cdot s \geq k^2 - 2ps \geq 2ps
     \]

     \[
     \Longleftrightarrow 1 + 2 \sqrt{p^2 - 1} \cdot s + 2ps \geq k^2 \geq 4ps
     \]

   - Since \( \sqrt{p^2 - 1} < \sqrt{p^2} = p \), this yields:

     \[
     1 + 2p \cdot s + 2ps > k^2 \geq 4ps \Longleftrightarrow 1 + 4ps > k^2 \geq 4ps
     \]

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

     \[
     x = \frac{p + s}{2} + \frac{k}{2} = \frac{p}{2} + \frac{k}{2}
     \]

     \[
     y = \frac{p + s}{2} - \frac{k}{2} = \frac{p}{2} - \frac{k}{2}
     \]

   - Now,

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

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

     \[
     \Longleftrightarrow \left( \sqrt{p-1} + \sqrt{p+1} \right)^2 \leq \left( \sqrt{p + k} + \sqrt{p - k} \right)^2
     \]

     \[
     \Longleftrightarrow (p - 1) + (p + 1) + 2 \sqrt{p - 1} \cdot \sqrt{p + 1} \leq (p + k) + (p - k) + 2 \sqrt{p + k} \cdot \sqrt{p - k}
     \]

     \[
     \Longleftrightarrow 2p + 2 \sqrt{p^2 - 1} \leq 2p + 2 \sqrt{p^2 - k^2} \Longleftrightarrow 1 \geq k^2
     \]

   - Since \( k \) is an integer and \( k \leq 0 \), this yields either \( k = 0 \) or \( k = -1 \). For \( k = 0 \), we have \( x = \frac{p}{2} + \frac{k}{2} = \frac{p}{2} \), which is impossible since \( p \) is odd and \( x \) must be an integer. Thus, \( k = -1 \), so that:

     \[
     x = \frac{p}{2} + \frac{k}{2} = \frac{p - 1}{2}
     \]

     \[
     y = \frac{p}{2} - \frac{k}{2} = \frac{p + 1}{2}
     \]

     Thus, Assertion 2 is proven.

4. **Assertion 3**: Let \( x \) and \( y \) be positive integers such that \( x \leq y \) and such that \( \sqrt{2p} - \sqrt{x} - \sqrt{y} \) is nonnegative. If \( (x, y) \neq \left( \frac{p-1}{2}, \frac{p+1}{2} \right) \), then:

   \[
   \sqrt{2p} - \sqrt{x} - \sqrt{y} > \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}}
   \]

   (In fact, otherwise we would have \( \sqrt{2p} - \sqrt{x} - \sqrt{y} \leq \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}} \), and thus Assertion 2 would yield \( x = \frac{p-1}{2} \) and \( y = \frac{p+1}{2} \), which contradicts \( (x, y) \neq \left( \frac{p-1}{2}, \frac{p+1}{2} \right) \).)

5. **Conclusion**: The number \( \sqrt{2p} - \sqrt{\frac{p-1}{2}} - \sqrt{\frac{p+1}{2}} \) is indeed the smallest possible nonnegative value of the term \( \sqrt{2p} - \sqrt{x} - \sqrt{y} \) for positive integers \( x \) and \( y \) with \( x \leq y \), and this value is achieved only for \( x = \frac{p-1}{2} \) and \( y = \frac{p+1}{2} \).

\(\blacksquare\)

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