Radical Rationals!

Under radical must be a perfect square. Thus for co-prime positive integers $p$ and $q$ $\frac{a+b}{ab}=\frac{p^2}{q^2}$ We get $(a+b)q^2=abp^2$ . We have that $\gcd(a+b, ab)=1$ because $a$ and $b$ are co-prime. Thus $a+b \mid p^2$ and $p^2 \mid a+b$ . Hence $a+b=p^2$ and similarly $ab=q^2$ .

Now, product of $a$ and $b$ is a perfect square. Thus both of them are perfect squares Let $a=m^2$ and $b=n^2$ . Hence $m^2+n^2=p^2$ and $(m, n, p)$ are primitive Pythagorean triples. That is $m=x^2-y^2$ and $n=2xy$ , where $x$ and $y$ are co-prime positive integers with different parity.

$a$ and $b$ are less than 1000, thus $x^2-y^2<\sqrt{1000}$ and $2xy<\sqrt{1000}$ . So we need to find the pairs $(x, y)$ for which $x^2-y^2 \le 31$ and $xy \le 15$ . It's easy to see that the pairs are $(x, y)=(5, 2), (4, 3), (4, 1), (3, 2), (2, 1)$ .

Then from $a=(x^2-y^2)^2$ and $b=4x^2y^2$ , we get $\sum_{a, b} \sqrt{\frac1a+\frac1b}=\sum_{x, y} \sqrt{\frac{1}{(x^2-y^2)^2}+\frac{1}{4x^2y^2}}=\sum_{x, y} \left( \frac{x^2+y^2}{2xy(x^2-y^2)}\right) =\frac{29}{420}+\frac{25}{168}+\frac{17}{120}+\frac{13}{60}+\frac{5}{12}=\frac{139}{140}$ Thus $\boxed{n=139}$ is the answer to this question.