Mathematica:
$$\int_{0}^{\infty} x \;K_{a}(nx) \; K_{b}(kx) \; dx=\frac{1}{2n^2}(k n)^{-b} $$
$$\qquad\times \left[n^{2 b} \Gamma (b) \Gamma \left(-\frac{a}{2}-\frac{b}{2}+1\right) \Gamma \left(\tfrac{1}{2} (a-b+2)\right) \, _2F_1\left(\tfrac{1}{2} (-a-b+2),\tfrac{1}{2} (a-b+2);1-b;\frac{k^2}{n^2}\right)\right.$$
$$\qquad\left.+\,k^{2 b} \Gamma (-b) \Gamma \left(\tfrac{1}{2} (-a+b+2)\right) \Gamma \left(\tfrac{1}{2} (a+b+2)\right) \, _2F_1\left(\tfrac{1}{2} (-a+b+2),\tfrac{1}{2} (a+b+2);b+1;\frac{k^2}{n^2}\right)\right]$$

for $-2<{\rm Re}(a-b)<2$, $-2<{\rm Re}(a+b)<2$

if $k=n$ this simplifies to

$$\int_{0}^{\infty} x \;K_{a}(nx) \; K_{b}(nx)=\frac{\pi ^2 (a^2-b^2)}{4 n^2 (\cos \pi b-\cos \pi a)}$$

there do not seem to be further simplifications if $k\neq n$