Let the required natural number be $x$.

According to the given condition,
$x+x^{2}=156$
$\Rightarrow x^{2}+x-156=0$
$\Rightarrow x^{2}+13 x-12 x-156=0$
$\Rightarrow x(x+13)-12(x+13)=0$
$\Rightarrow(x+13)(x-12)=0$
$\Rightarrow x+13=0$ or $x-12=0$
$\Rightarrow x=-13$ or $x=12$ $\therefore x=12 \quad(x$ cannot be negative)

Hence, the required natural number is $12 .$