Suppose that $A$ and $B$ were rational numbers of rank 3 less than $\frac{1}{4}$, and let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ be positive integers so that $A=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}$ and $B=\frac{1}{b_{1}}+\frac{1}{b_{2}}+\frac{1}{b_{3}}$ are the expressions for $A$ and $B$ as stated in the problem. If $b_{1}<a_{1}$ then $A<\frac{1}{a_{1}-1} \leq \frac{1}{b_{1}}<B$. In other words, of all the rationals less than $\frac{1}{4}$ with rank 3, those that have $a_{1}=5$ are greater than those that have $a_{1}=6,7,8, \ldots$ Therefore we can "build" $q$ greedily, adding the largest unit fraction that keeps $q$ less than $\frac{1}{4}$: $\frac{1}{5}$ is the largest unit fraction less than $\frac{1}{4}$, hence $a_{1}=5$; $\frac{1}{27}$ is the largest unit fraction less than $\frac{1}{4}-\frac{1}{5}$, hence $a_{2}=21$; $\frac{1}{421}$ is the largest unit fraction less than $\frac{1}{4}-\frac{1}{5}-\frac{1}{21}$, hence $a_{3}=421$.