Find all triples $(x,y,z)$ of positive rationals such that $x + y + z$, $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$ and $xyz$ are all integers.
1. **Define the given conditions:**
   We are given that \(x + y + z\), \(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\), and \(xyz\) are all integers. Let:
   \[
   a = x + y + z, \quad b = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}, \quad c = xyz
   \]
   Since \(x, y, z\) are positive rationals, we can assume \(x = \frac{p}{q}\), \(y = \frac{r}{s}\), \(z = \frac{u}{v}\) where \(p, q, r, s, u, v\) are positive integers.

2. **Express the polynomial:**
   The polynomial whose roots are \(x, y, z\) is:
   \[
   t^3 - at^2 + bct - c = 0
   \]
   By Vieta's formulas, we know:
   \[
   x + y + z = a, \quad xy + xz + yz = bc, \quad xyz = c
   \]

3. **Rational Root Theorem:**
   Since \(x, y, z\) are rational, by the Rational Root Theorem, they must be integers. This is because the coefficients of the polynomial are integers, and the Rational Root Theorem states that any rational solution \( \frac{p}{q} \) must have \( p \) as a factor of the constant term and \( q \) as a factor of the leading coefficient. Here, the leading coefficient is 1, so \( q = 1 \), implying \( x, y, z \) are integers.

4. **Check integer solutions:**
   We need to find integer solutions \(x, y, z\) such that:
   \[
   \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = b
   \]
   is also an integer. We can test small integer values for \(x, y, z\):

   - For \(x = 1, y = 1, z = 1\):
     \[
     x + y + z = 3, \quad \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 3, \quad xyz = 1
     \]
     This satisfies all conditions.

   - For \(x = 1, y = 2, z = 2\):
     \[
     x + y + z = 5, \quad \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1 + \dfrac{1}{2} + \dfrac{1}{2} = 2, \quad xyz = 4
     \]
     This satisfies all conditions.

   - For \(x = 2, y = 3, z = 6\):
     \[
     x + y + z = 11, \quad \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = 1, \quad xyz = 36
     \]
     This satisfies all conditions.

   - For \(x = 2, y = 4, z = 4\):
     \[
     x + y + z = 10, \quad \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} = 1, \quad xyz = 32
     \]
     This satisfies all conditions.

5. **Conclusion:**
   The integer solutions that satisfy all the given conditions are:
   \[
   \{x, y, z\} \in \{\{1, 1, 1\}, \{1, 2, 2\}, \{2, 3, 6\}, \{2, 4, 4\}\}
   \]

The final answer is \( \boxed{ \{ \{1, 1, 1\}, \{1, 2, 2\}, \{2, 3, 6\}, \{2, 4, 4\} \} } \)