Let the numbers in the pair be represented as reduced fractions $\frac{a}{b}$ and $\frac{c}{d}$. For
$$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$
to be an integer, we must have
$$ad + bc = k \cdot bd \quad (2)$$
with $k$ being some integer. By writing the equality (2) as $bc = (kb - a) \cdot d$ and taking into account that $c$ and $d$ are relatively prime, we see that $b$ is divisible by $d$. By writing the same equality (2) in a form $ad = (kd - c) \cdot b$ and taking into account that $a$ and $b$ are relatively prime, we see that $d$ is divisible by $b$. Therefore $b = d$.
For also $\frac{b}{a} + \frac{d}{c}$ to be an integer, we analogously must have $a = c$. Therefore $\frac{a}{b} = \frac{c}{d}$.
Consequently $\frac{a}{b} + \frac{c}{d} = \frac{2a}{b}$ and $\frac{b}{a} + \frac{d}{c} = \frac{2b}{a}$ are both integers. If $a = b$, then $\frac{a}{b} = 1$. If $a < b$, then $\frac{2a}{b} < 2$, implying $\frac{2a}{b} = 1$ and $\frac{a}{b} = \frac{1}{2}$ as the only possibility. If $a > b$, then similarly $\frac{a}{b} = 2$.

Let the numbers in the pair be represented as reduced fractions $\frac{a}{b}$ and $\frac{c}{d}$, and let $k = \frac{a}{b} + \frac{c}{d}$ and $l = \frac{b}{a} + \frac{d}{c}$. Then $\frac{a}{b} = k - \frac{c}{d} = \frac{kd - c}{d}$ and $\frac{b}{a} = l - \frac{d}{c} = \frac{lc - d}{c}$, whence $\frac{a}{b} = \frac{c}{lc - d}$. Therefore $\frac{kd - c}{d} = \frac{c}{lc - d}$, or $cd = (kd - c)(lc - d)$. The resulting equation is equivalent to $l \cdot c^2 - kld \cdot c + kd^2 = 0$, giving
$$c = \frac{kld \pm \sqrt{k^2 l^2 d^2 - 4kld^2}}{2l} = d \cdot \frac{kl \pm \sqrt{k^2 l^2 - 4kl}}{2l} \quad (3)$$
For $c$ to be an integer, we must have $k^2 l^2 - 4kl = n^2$ where $n$ is an integer. Now
$$(kl - 2)^2 = n^2 + 4.$$
Therefore $n^2 + 4$ must also be a square of an integer. This is only possible when $n = 0$ – therefore $kl = 0$ or $kl = 4$. The first option is not possible, because $k$ and $l$ are the sums
of positive real numbers. The second option gives three possible cases: $(k,l)$ can either be $(1,4), (2,2)$ or $(4,1)$. It remains to find all possible values of $\frac{c}{d}$ from (3) and calculate $\frac{a}{b} = k - \frac{c}{d}$.

For also $\frac{b}{a} + \frac{d}{c}$ to be an integer, we analogously must have $a = c$. Therefore $\frac{a}{b} = \frac{c}{d}$.
Consequently $\frac{a}{b} + \frac{c}{d} = \frac{2a}{b}$ and $\frac{b}{a} + \frac{d}{c} = \frac{2b}{a}$ are both integers. If $a = b$, then $\frac{a}{b} = 1$. If $a < b$, then $\frac{2a}{b} < 2$, implying $\frac{2a}{b} = 1$ and $\frac{a}{b} = \frac{1}{2}$ as the only possibility. If $a > b$, then similarly $\frac{a}{b} = 2$.