1074*. Find all such rational numbers $x$ and $y$ for which the numbers $x+\frac{1}{y}$ and $y+\frac{1}{x}$ are natural.
$\triangle$ Since the numbers

$$
x+\frac{1}{y}=\frac{x y+1}{y} \text { and } y+\frac{1}{x}=\frac{x y+1}{x}
$$

are positive, the number

$$
\frac{x y+1}{y} \cdot \frac{x y+1}{x}=\frac{(x y+1)^{2}}{x y}
$$

is also positive, which implies that $x y$ is positive. But then the number $x y+1$ is positive, and consequently, the numbers $x$ and $y$ are positive.

Let

$$
x=\frac{a}{b}, \quad y=\frac{k}{n}
$$

where $a, b, k, n$ are natural numbers, and the fractions $\frac{a}{b}$ and $\frac{k}{n}$ are irreducible.

We get:

$$
\begin{aligned}
& \frac{x y+1}{y}=\left[\frac{a k}{b n}+1\right]: \frac{k}{n}=\frac{a k+b n}{b k} \\
& \frac{x y+1}{x}=\left[\frac{a k}{b n}+1\right]: \frac{k}{n}=\frac{a k+b n}{a n}
\end{aligned}
$$

These fractions must be equal to natural numbers. Let's consider the first one.

Since the sum $a k+b n$ is divisible by $b$, then $a k$ is divisible by $b$. But the numbers $a$ and $b$ are coprime, so $k$ is divisible by $b$.

Similarly, since the sum $a k+b n$ is divisible by $k$, then $b n$ is divisible by $k$, which implies that $b$ is divisible by $k$.

From the divisibilities $k: b$ and $b: k$, it follows that $k=b$.

Similarly, from the fact that the second fraction is equal to a natural number, we deduce that $n=a$.

Then these fractions take the following form:

$$
\frac{a k+b n}{b k}=\frac{a b+b a}{b^{2}}=\frac{2 a}{b}, \quad \frac{a k+b n}{a n}=\frac{a b+b a}{a^{2}}=\frac{2 b}{a}
$$

Since the obtained fractions $\frac{2 a}{b}$ and $\frac{2 b}{a}$ are equal to natural numbers, and the numbers $a$ and $b$ are coprime, then $2: a$ and $2: b$. Therefore, the numbers $a$ and $b$ can only be 1 or 2.

Answer: 1, $1 ; 2,1 / 2 ; 1 / 2,2$.