Find all triplets of positive rational numbers $(m,n,p)$ such that the numbers $m+\frac 1{np}$, $n+\frac 1{pm}$, $p+\frac 1{mn}$ are integers.

[i]Valentin Vornicu, Romania[/i]
1. Let \( m, n, p \) be positive rational numbers. Define \( t = mnp \). We are given that \( m + \frac{1}{np} \), \( n + \frac{1}{pm} \), and \( p + \frac{1}{mn} \) are integers. Let's denote these integers by \( a, b, \) and \( c \) respectively. Thus, we have:
   \[
   a = m + \frac{1}{np}, \quad b = n + \frac{1}{pm}, \quad c = p + \frac{1}{mn}
   \]

2. Multiply all three equations together:
   \[
   \left(m + \frac{1}{np}\right)\left(n + \frac{1}{pm}\right)\left(p + \frac{1}{mn}\right) = abc
   \]
   Substituting \( t = mnp \), we get:
   \[
   \left(m + \frac{1}{np}\right)\left(n + \frac{1}{pm}\right)\left(p + \frac{1}{mn}\right) = \left(m + \frac{1}{np}\right)\left(n + \frac{1}{pm}\right)\left(p + \frac{1}{mn}\right) = mnp \left(1 + \frac{1}{mnp}\right)^3 = t \left(1 + \frac{1}{t}\right)^3
   \]
   Simplifying the right-hand side:
   \[
   t \left(1 + \frac{1}{t}\right)^3 = t \left(\frac{t+1}{t}\right)^3 = t \cdot \frac{(t+1)^3}{t^3} = \frac{(t+1)^3}{t^2}
   \]
   Therefore:
   \[
   abc = \frac{(t+1)^3}{t^2}
   \]

3. Since \( abc \) is an integer, \(\frac{(t+1)^3}{t^2}\) must also be an integer. This implies that \( t \) must be an integer because \( t \) is a rational number and \( t \) must divide \((t+1)^3\). 

4. The greatest common divisor of \( t \) and \( t+1 \) is 1, i.e., \(\gcd(t, t+1) = 1\). Therefore, \( t \) must be 1 because if \( t \) were greater than 1, some prime would divide the denominator but not the numerator.

5. Thus, \( t = 1 \). Substituting \( t = 1 \) back into the equations, we get:
   \[
   mnp = 1
   \]
   and the integers \( a, b, \) and \( c \) become:
   \[
   a = m + \frac{1}{np} = m + \frac{1}{\frac{1}{m}} = 2m
   \]
   \[
   b = n + \frac{1}{pm} = n + \frac{1}{\frac{1}{n}} = 2n
   \]
   \[
   c = p + \frac{1}{mn} = p + \frac{1}{\frac{1}{p}} = 2p
   \]

6. Since \( 2m, 2n, 2p \) are integers and their product is:
   \[
   8mnp = 8 \cdot 1 = 8
   \]
   We need to find the integer solutions to:
   \[
   2m \cdot 2n \cdot 2p = 8
   \]
   The possible sets of integers \(\{2m, 2n, 2p\}\) that satisfy this equation are:
   \[
   \{1, 1, 8\}, \{1, 2, 4\}, \{2, 2, 2\}
   \]

7. Therefore, the corresponding rational solutions \((m, n, p)\) are:
   \[
   (m, n, p) = \left(\frac{1}{2}, \frac{1}{2}, 4\right), \left(\frac{1}{2}, 1, 2\right), (1, 1, 1)
   \]
   and their permutations.

The final answer is \(\boxed{\left(\frac{1}{2}, \frac{1}{2}, 4\right), \left(\frac{1}{2}, 1, 2\right), (1, 1, 1)}\) and permutations.