<s>[INST] positive fractions, denominator 4, difference equals quotient (4,2) are the only positive integers whose difference is equal to their quotient. Find the sum of two positive fractions, in their lowest terms, whose denominators are 4 that also share this same property.
 [/INST] I prove your first sentence below:
If $x, y\in\mathbb Z$ satisfy $x-y=x/y,$ then we have :
$$x(y-1)=y^2.$$
Thus, if a prime $p$ divides $y-1,$ then $p$ divides $y$ as well, a contradiction as $y$ and $y-1$ are coprime. Hence no prime divides $y-1,$ i.e. $y=2.$ Therefore $x(2-1)=4.$  
As to the second sentence, if $(x-y)/4=x/y,$ then we have:
$$x(y-4)=y^2.$$
Since any common divisor of $y$ and $y-4$ must divide $4,$ denoting $\gcd(y,y-4)=g,$ we consider three cases:
I. $g=1.$
We have already shown that, in this case, $y-4=1,$ i.e. $y=5$ and $x=25.$
II. $g=2.$
In this case $y-4$ cannot have any odd prime divisor, and $4$ cannot divide $y-4$ either, thus $y-4=2,$ i.e. $y=6$ and $x=18.$
III. $g=4.$
Again no common odd prime divisor of $y$ and $y-4$ can occur, while $4$ divides $y-4.$ Hence $y-4=4,$ i.e. $y=8,$ and $x=16.$
So you know all the positive fractions with the required property, and it remains to calculate the sum of two of them, which I left to you.
Hope this helps.
Edit:
The fractions are supposed to be in the lowest terms, thus the cases II. and III. do not count at all. Hence the only pair of fractions with the required property is $(\frac{x}{4},\frac{y}{4})=(\frac{25}{4},\frac{5}{4}),$ in which case the sum is $\frac{15}{2}.$
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