Four. (25 points) Do there exist two positive integers $a, b$, satisfying $a \leqslant b$, such that the equation in $x$
$$
x^{2}-a b x+a+b=0
$$

has two integer solutions? If they exist, find all such positive integers $a, b$; if not, provide an explanation.
Therefore, $x_{1}, x_{2}$ are both positive integers, and
$$
x_{1} x_{2}-x_{1}-x_{2}=a+b-a b \text {, }
$$

i.e., $\square$
$$
\begin{array}{l}
\left(x_{1}-1\right)\left(x_{2}-1\right) \\
=-(a-1)(b-1)+2 .
\end{array}
$$

By $\left(x_{1}-1\right)\left(x_{2}-1\right) \geqslant 0$, we know
$$
0 \leqslant M \leqslant 2(M=(a-1)(b-1)) \text {. }
$$

Thus, $M=0$ or 1 or 2.
(1) When $M=2$, we have $a-1=1, b-1=2$, i.e., $a=2, b=3$. From equation (1), we get $x_{1}=1, x_{2}=5$.
(2) When $M=1$, we have $a-1=1, b-1=1$, i.e., $a=2, b=2$. From equation (1), we get $\dot{x}_{1}=x_{2}=2$, which meets the conditions.
(3) When $M=0$, we have $a-1=0$, i.e., $a=1$. From equation (1), we get $x_{1}=2, x_{2}=3$. Thus, $x_{1} x_{2}=6=a+b$ $=1+b$. Hence, $b=5$.
In summary, the positive integers that meet the conditions are
$$
a=2, b=3 ; a=b=2 ; a=1, b=5 \text {. }
$$

Four, assuming that the positive integers $a, b$ meet the conditions, such that the equation $x^{2}-a b x+a+b=0$ has two integer solutions $x_{1}, x_{2}$ (assuming $x_{1} \leqslant x_{2}$). By the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=a b>0, x_{1} x_{2}=a+b>0 .
$$