The sum of two numbers is 16.

Let one numbers be $x$ then other $(16-x)$.

Then according to question

$\frac{1}{x}+\frac{1}{(16-x)}=\frac{1}{3}$

$\frac{16}{\left(16 x-x^{2}\right)}=\frac{1}{3}$

By cross multiplication

$16 x-x^{2}=48$

$x^{2}-16 x+48=0$

$x^{2}-12 x-4 x-48=0$

$x(x-12)-4(x-12)=0$

$(x-12)(x-4)=0$

$(x-12)=0$

$x=12$

Or

$(x-4)=0$

$x=4$

Since, x being a number,

Therefore,

When $x=12$ then

$16-x=16-12$

$=4$

Thus, two consecutive number be either 4,12