Problem 1235

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.