Determine the ordered systems $(x,y,z)$ of positive rational numbers for which $x+\frac{1}{y},y+\frac{1}{z}$ and $z+\frac{1}{x}$ are integers.
1. Let \( x = \frac{p}{q} \), \( y = \frac{r}{s} \), and \( z = \frac{t}{u} \) where \( (p, q) \), \( (r, s) \), and \( (t, u) \) are pairs of coprime positive integers. We need to find the ordered systems \((x, y, z)\) such that \( x + \frac{1}{y} \), \( y + \frac{1}{z} \), and \( z + \frac{1}{x} \) are integers.

2. Consider the expression \( x + \frac{1}{y} \):
   \[
   x + \frac{1}{y} = \frac{p}{q} + \frac{s}{r} = \frac{pr + qs}{qr}
   \]
   For this to be an integer, \( \frac{pr + qs}{qr} \) must be an integer, implying \( qr \mid (pr + qs) \).

3. Since \( (p, q) \) and \( (r, s) \) are coprime pairs, \( q \mid pr \) and \( r \mid qs \). Given that \( q \) and \( r \) are coprime, it follows that \( q \mid s \) and \( r \mid q \). Therefore, \( q = r \).

4. Similarly, consider the expression \( y + \frac{1}{z} \):
   \[
   y + \frac{1}{z} = \frac{r}{s} + \frac{u}{t} = \frac{rt + su}{st}
   \]
   For this to be an integer, \( \frac{rt + su}{st} \) must be an integer, implying \( st \mid (rt + su) \).

5. Since \( (r, s) \) and \( (t, u) \) are coprime pairs, \( s \mid rt \) and \( t \mid su \). Given that \( s \) and \( t \) are coprime, it follows that \( s \mid u \) and \( t \mid s \). Therefore, \( s = t \).

6. Finally, consider the expression \( z + \frac{1}{x} \):
   \[
   z + \frac{1}{x} = \frac{t}{u} + \frac{q}{p} = \frac{tp + qu}{up}
   \]
   For this to be an integer, \( \frac{tp + qu}{up} \) must be an integer, implying \( up \mid (tp + qu) \).

7. Since \( (t, u) \) and \( (p, q) \) are coprime pairs, \( u \mid tp \) and \( p \mid qu \). Given that \( u \) and \( p \) are coprime, it follows that \( u \mid q \) and \( p \mid u \). Therefore, \( u = p \).

8. From the above steps, we have \( q = r \), \( s = t \), and \( u = p \). This implies that \( x = \frac{p}{q} \), \( y = \frac{q}{s} \), and \( z = \frac{s}{p} \).

9. We need to find the values of \( p, q, s \) such that \( pqs \mid (p + q + s) \). Given that \( p, q, s \) are positive integers, we can test small values to find solutions.

10. Testing small values, we find the following solutions:
   - \( (p, q, s) = (1, 1, 1) \)
   - \( (p, q, s) = (1, 1, 2) \)
   - \( (p, q, s) = (1, 2, 3) \)

11. Converting these back to the original variables \( x, y, z \), we get the following ordered systems:
   - \( (1, 1, 1) \)
   - \( (1, \frac{1}{2}, 2) \)
   - \( (\frac{1}{2}, 2, 1) \)
   - \( (2, 1, \frac{1}{2}) \)
   - \( (\frac{1}{2}, \frac{2}{3}, 3) \)
   - \( (3, \frac{1}{2}, \frac{2}{3}) \)
   - \( (\frac{2}{3}, 3, \frac{1}{2}) \)
   - \( (\frac{1}{3}, \frac{3}{2}, 2) \)
   - \( (2, \frac{1}{3}, \frac{3}{2}) \)
   - \( (\frac{3}{2}, 2, \frac{1}{3}) \)

The final answer is \( \boxed{ (1, 1, 1) } \), \((1, \frac{1}{2}, 2)\), \((\frac{1}{2}, 2, 1)\), \((2, 1, \frac{1}{2})\), \((\frac{1}{2}, \frac{2}{3}, 3)\), \((3, \frac{1}{2}, \frac{2}{3})\), \((\frac{2}{3}, 3, \frac{1}{2})\), \((\frac{1}{3}, \frac{3}{2}, 2)\), \((2, \frac{1}{3}, \frac{3}{2})\), \((\frac{3}{2}, 2, \frac{1}{3})\).