$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ where $a,b,c\in\mathbb{N}$ such that $1\leq a\leq b\leq c$.
$\frac{4}{3n}\leq\frac{1}{a}<\frac{4}{n}$, so $\frac{n}{4}<a\leq\frac{3n}{4}$
$\frac{\frac{4}{n}-\frac{1}{a}}{2}\leq\frac{1}{b}<\frac{4}{n}-\frac{1}{a}$, so $\frac{1}{\frac{4}{n}-\frac{1}{a}}<b\leq\frac{2}{\frac{4}{n}-\frac{1}{a}}$
$\Rightarrow na<(4a-n)b\leq 2na$
$\frac{1}{c}=\frac{4}{n}-\frac{1}{a}-\frac{1}{b}$, so $c=\frac{1}{\frac{4}{n}-\frac{1}{a}-\frac{1}{b}} \Rightarrow c=\frac{nab}{4ab-nb-na}$