Determine all integer numbers $x$ and $y$ such that:

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{19}
$$
The equation is equivalent to $x y=19(x+y)$. Since we are looking for integer solutions and 19 is a prime number, this equality implies that $x$ or $y$ must be divisible by 19. As the equation is symmetric with respect to the variables $x$ and $y$, we can assume that $x$ is divisible by 19. That is, $x=19 k$ for some integer value of $k$. Under this condition, we have:

$$
x y=19(x+y) \quad \Leftrightarrow \quad k y=19 k+y
$$

From this equality, we conclude that $19 k+y$ is divisible by $k$. Since $19 k$ is already divisible by $k$, we conclude that $y$ is divisible by $k$ (prove this). That is, $y=k m$ for some integer value of $m$. Thus,

$$
k y=19 k+y \Rightarrow k^{2} m=19 k+k m \Rightarrow k m-m=19 \Rightarrow m(k-1)=19
$$

Since $m$ and $k$ are integers and 19 is a prime number, there are only 4 possibilities for the equality $m(k-1)=19$:

- $m=19$ and $k-1=1$. This implies that $x=38$ and $y=38$;
- $m=-19$ and $k-1=-1$. This implies that $x=0$ and $y=0$, which is not possible, as in the original equation, $x \neq 0$ and $y \neq 0$;
- $m=1$ and $k-1=19$. This implies that $x=380$ and $y=20$;
- $m=-1$ and $k-1=-19$. This implies that $x=-342$ and $y=18$.

Thus, we obtain the following possibilities for the pairs of integers $(x, y)$ that are solutions to the given equation:

$$
(38,38) ;(380,20) ;(20,380) ;(-342,18) ;(18,-342)
$$