Question 191638
Let n and n+1 be the two consecutive numbers.

Then {{{1/n -1/(n+1)=1/12}}}

Solving for n, we have

{{{12(n+1 -n) = n(n+1)}}}
{{{12=n^2+n}}}
{{{n^2+n-12=0}}}
{{{(n+4)(n-3)=0}}}
so n= -4, or n=3
If n= -4, the two consecutive numbers are -4, -3.
If n = 3, the two consecutive numbers are 3, 4.