Let the required natural numbers be $x$ and $(9,-x)$.

According to the given condition,

$\begin{array}{l}
\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2} \\
\Rightarrow \frac{9-x+x}{x(9-x)}=\frac{1}{2} \\
\Rightarrow \frac{9}{9 x-x^{2}}=\frac{1}{2} \\
\Rightarrow 9 x-x^{2}=18 \\
\Rightarrow x^{2}-9 x+18=0 \\
\Rightarrow x^{2}-6 x-3 x+18=0 \\
\Rightarrow x(x-6)-3(x-6)=0 \\
\Rightarrow x-3=0 \text { or } x-6=0 \\
\Rightarrow x=3 \text { or } x=6
\end{array}$

When $x=3$
$\begin{array}{l}
9-x=9-3=6 \\
\text { When } x=6 \\
9-x=9-6=3
\end{array}$

Hence, the required natural numbers are 3 and $6 .$