The sum of two natural numbers is 9 and the sum of their reciprocals is

Let the required natural numbers be x and (9 − x).

According to the given condition,

$\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}$

$\Rightarrow \frac{9-x+x}{x(9-x)}=\frac{1}{2}$

$\Rightarrow \frac{9}{9 x-x^{2}}=\frac{1}{2}$

$\Rightarrow 9 x-x^{2}=18$

$\Rightarrow x^{2}-9 x+18=0$

$\Rightarrow x^{2}-6 x-3 x+18=0$

$\Rightarrow x(x-6)-3(x-6)=0$

$\Rightarrow(x-3)(x-6)=0$

$\Rightarrow x-3=0$ or $x-6=0$

$\Rightarrow x=3$ or $x=6$

When x = 3,
9 − x = 9 − 3 = 6

When x = 6,
9 − x = 9 − 6 = 3

Hence, the required natural numbers are 3 and 6.