Determine all positive integers $n$ for which there are positive real numbers $x,y$   and  $z$  such that $\sqrt x +\sqrt y +\sqrt z=1$ and $\sqrt{x+n} +\sqrt{y+n} +\sqrt{z+n}$ is an integer. 
1. **Given Conditions:**
   - We need to find all positive integers \( n \) such that there exist positive real numbers \( x, y, z \) satisfying:
     \[
     \sqrt{x} + \sqrt{y} + \sqrt{z} = 1
     \]
     and
     \[
     \sqrt{x+n} + \sqrt{y+n} + \sqrt{z+n}
     \]
     is an integer.

2. **Applying Jensen's Inequality:**
   - By Jensen's inequality for the concave function \( f(t) = \sqrt{t} \), we have:
     \[
     \sqrt{x+n} + \sqrt{y+n} + \sqrt{z+n} \geq 3 \sqrt{\frac{x+n + y+n + z+n}{3}} = 3 \sqrt{\frac{x+y+z + 3n}{3}} = 3 \sqrt{\frac{1 + 3n}{3}} = \sqrt{9n + 1}
     \]

3. **Upper Bound Estimation:**
   - We also need an upper bound for \( \sqrt{x+n} + \sqrt{y+n} + \sqrt{z+n} \). Using the inequality \( \sqrt{a} + \sqrt{b} \leq \sqrt{2(a+b)} \), we get:
     \[
     \sqrt{x+n} + \sqrt{y+n} + \sqrt{z+n} \leq \sqrt{2(x+n) + 2(y+n)} + \sqrt{z+n} \leq \sqrt{2(x+y+2n)} + \sqrt{z+n}
     \]
     Since \( x + y + z = 1 \), we can further simplify:
     \[
     \sqrt{x+n} + \sqrt{y+n} + \sqrt{z+n} \leq 2\sqrt{n} + \sqrt{n+1}
     \]

4. **Combining Inequalities:**
   - We now have:
     \[
     \sqrt{9n + 1} \leq \sqrt{x+n} + \sqrt{y+n} + \sqrt{z+n} \leq 2\sqrt{n} + \sqrt{n+1}
     \]

5. **Integer Condition:**
   - Let \( \sqrt{x+n} + \sqrt{y+n} + \sqrt{z+n} = m \), where \( m \) is an integer. Then:
     \[
     \sqrt{9n + 1} \leq m \leq 2\sqrt{n} + \sqrt{n+1}
     \]
     Squaring the inequalities, we get:
     \[
     9n + 1 \leq m^2 \leq (2\sqrt{n} + \sqrt{n+1})^2
     \]

6. **Simplifying the Upper Bound:**
   - Expanding the upper bound:
     \[
     (2\sqrt{n} + \sqrt{n+1})^2 = 4n + 4\sqrt{n(n+1)} + (n+1) = 5n + 1 + 4\sqrt{n(n+1)}
     \]
     Since \( 4\sqrt{n(n+1)} \) is always positive, we have:
     \[
     m^2 \leq 5n + 1 + 4\sqrt{n(n+1)}
     \]
     For large \( n \), \( 4\sqrt{n(n+1)} \approx 4n \), so:
     \[
     m^2 \leq 9n + 1 + \epsilon
     \]
     where \( \epsilon \) is a small positive number.

7. **Perfect Square Condition:**
   - For \( m^2 \) to be an integer, \( 9n + 1 \) must be a perfect square. Let \( 9n + 1 = k^2 \) for some integer \( k \). Then:
     \[
     9n = k^2 - 1
     \]
     \[
     n = \frac{k^2 - 1}{9}
     \]
     For \( n \) to be an integer, \( k^2 - 1 \) must be divisible by 9. This implies \( k \equiv \pm 1 \pmod{9} \).

8. **Conclusion:**
   - Therefore, the positive integers \( n \) that satisfy the given conditions are those for which \( 9n + 1 \) is a perfect square.

The final answer is \( \boxed{ n = \frac{k^2 - 1}{9} } \) where \( k \equiv \pm 1 \pmod{9} \).