The difference of two numbers is 4.

Let one numbers be $x$ then other $(x+4)$.

Then according to question

$\frac{1}{x}-\frac{1}{(x+4)}=\frac{4}{21}$

$\frac{4}{\left(x^{2}+4 x\right)}=\frac{4}{21}$

By cross multiplication

$4 x^{2}+16 x=84$

$4 x^{2}+16 x-84=0$

$4\left(x^{2}+4 x-21\right)=0$

$\left(x^{2}+4 x-21\right)=0$

$x^{2}+7 x-3 x-21=0$

$x(x+7)-3(x+7)=0$

$(x+7)(x-3)=0$

$(x+7)=0$

$x=-7$

Or

$(x-3)=0$

$x=3$

Since, x being a number,

Therefore,

When $x=-7$ then

$x+4=-7+4$

$=-3$

And when $x=3$ then

$x+4=3+4$

$=7$

Thus, two consecutive number be either 7,3 or $-7,-3$