Halved long division

Let us assign the unfilled boxes as below $\LARGE{ \begin{array}{rll} \phantom{0}\ \boxed{6} && \\[-2pt] \boxed{\phantom{0}x} \ \boxed{\phantom{0}y}\ \enclose{longdiv}{\boxed{\phantom{0}a} \ \boxed{7}}\kern-.2ex \\[-2pt] \underline{\boxed{\phantom{0}b} \ \boxed{8}} && \\[-2pt] \boxed{9} \end{array} }$ From above long division we can draw out $\begin{aligned}(10a+7)-(10b+8) =9 \Rightarrow a-b= 1 \end{aligned}$ We conclude that the difference of $a$ and $b$ is 1 which makes us clear that they are co-prime( moreover they are consecutive integers) to each other where $a> b$ .

Also we can have $\begin{aligned}6\times xy = 10b +8 \Rightarrow 5b+4 =3\times xy \end{aligned}$ Adding $1$ on both sides we get $\begin{aligned} &5(b+1) = 3\times xy +1 \end{aligned}$ or $\begin{aligned} &5a = 3\times xy +1 \\& 5a = 30x+3y +1 \end{aligned}$ The range of x is $0<x \leq 3$ .If we input either $x= 2$ or $x=3$ then $a$ will be two digits number(which is not as $a$ is single digit). So the appropriate value for $x$ is $1$ . Entering $x=1$ equation becomes $\begin{aligned}& 5a = 31 +3y\end{aligned}$ This equation will be true and $a$ will be a single digit if $31+3y < 50$ for which the value for $y = 3$ . Then $\begin{aligned}a = \frac{31+3\times3 }{5} = 8 \phantom {cccccc}, b = 8-1 =7\end{aligned}$ Therefore the sum of $a+b+x+y = 8+7+1+3=\boxed{19}$ .

2nd Solution written using the hints provided by Sir @Pi Han Goh

Since $\begin{aligned}& xy < \frac{100}{6} , xy >10 \\& \implies 10 < xy < \frac{100}{6} \\& \implies 60 < 6xy < 100\\& \implies 60 < (10b +8) \ < 100\\& \implies 10 < \frac{10b+8}{6}< \frac{100}{6}\\& \implies 10 < \frac{10(b-1)+18}{6}<16\\& \implies 10 < \frac{10(b-1)}{6} +3 < 16 \\& \end{aligned}$ $\frac{10(b-1)} {6}$ will be an integer iff $b-1$ multiple integral of 6. So $\begin{aligned} & b-1 = 6n , n \in N \\& \Rightarrow b = 6n+1 \\& b = 7 , 13 , 19 ,\cdots 6n+1\end{aligned}$ Since $b$ is single digit number so only acceptable value of $b$ is $7$ . Hence the inequality above becomes $\begin{aligned}10< 13 <16\end{aligned}$

The long division is $\LARGE{ \begin{array}{rll} \phantom{0}\ \boxed{6} && \\[-2pt] \boxed{\phantom{0}1} \ \boxed{\phantom{0}3}\ \enclose{longdiv}{\boxed{\phantom{0}8} \ \boxed{7}}\kern-.2ex \\[-2pt] \underline{\boxed{\phantom{0}7} \ \boxed{8}} && \\[-2pt] \boxed{9} \end{array} }$