19. Given that $x, y$ are positive integers, and satisfy
$$
x y-(x+y)=2 p+q \text{, }
$$

where $p, q$ are the greatest common divisor and the least common multiple of $x$ and $y$, respectively. Find all such pairs $(x, y)(x \geqslant y)$.
19. From the problem, let $x=a p, y=b p(a, b$ be positive integers and $(a, b)=1, a \geqslant b)$. Then, $q=a b p$. At this point, the equation in the problem is $a p \cdot b p-(a p+b p)=2 p+a b p$.
By $p>0$, the above equation simplifies to
$(p-1) a b=a+b+2$.
Equation (1) shows that $0<p-1=\frac{1}{a}+\frac{1}{b}+\frac{2}{a b} \leqslant 4$.
Therefore, $p=2,3,4$ or 5.
(1) When $p=2$, equation (1) becomes $a b-a-b+1=3$, i.e., $(a-1)(b-1)=3=3 \times 1$. At this point, $(a, b)=2$, which does not meet the requirements.
(2) When $p=3$, equation (1) becomes $2 a b-a-b=2$, i.e., $(2 a-1)(2 b-1)=5 \times 1$.
(3) When $p=4$, equation (1) becomes $3 a b-a-b=2$, i.e., $(3 a-1)(3 b-1)=7 \times 1$. This system of equations has no integer solutions.
(4) When $p=5$, equation (1) becomes $4 a b-a-b=2$, i.e., $(4 a-1)(4 b-1)=9 \times 1=3 \times 3$. This system of equations has no integer solutions.
From $\left\{\begin{array}{l}4 a-1=3, \\ 4 b-1=3,\end{array}\right.$ we get $\left\{\begin{array}{l}a=1, \\ b=1 .\end{array}\right.$ Therefore, $\left\{\begin{array}{l}x=a p=5, \\ y=b p=5 .\end{array}\right.$
Note: If one case is missing or an extra solution that does not meet the problem's requirements is given, 2 points will be deducted.