The sum of a number and its reciprocal is 17/4.

Let a numbers be $x$ and its reciprocal is $\frac{1}{x}$

Then according to question

$x+\frac{1}{x}=\frac{17}{4}$

$\frac{x^{2}+1}{x}=\frac{17}{4}$

By cross multiplication

$4 x^{2}+4=17 x$

$4 x^{2}-17 x+4=0$

$4 x^{2}-17 x+4=0$

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

$x(4 x-1)-4(4 x-1)=0$

$(4 x-1)(x-4)=0$

$(4 x-1)=0$

$x=\frac{1}{4}$

Or

$(x-4)=0$

$x=4$

Thus, two consecutive number be either 4 or $\frac{1}{4}$