Question 24844
Try this Javier:  Let n be the first positive even integer and n+2 be the next consecutive positive even integer. You can write the problem as:

{{{1/n - 1/(n+2) = 1/60}}} Add the fractions on the left.  The common denominator is {{{n(n+2) = n^2+2n}}}
{{{((n+2) - n)/(n^2+2n) = 1/60}}} Simplify and solve for n.
{{{2/(n^2+2n) = 1/60}}} Multiply both sides of the equation by 60.
{{{120/(n^2+2n) = 1}}} Multiply both sides by {{{(n^2+2n)}}}
{{{120 = n^2 + 2n}}} Subtract 120 from both sides.
{{{n^2 + 2n - 120 = 0}}} Solve for n by factoring.
{{{(n - 10)(n + 12) = 0}}} Apply the zero products principle.
{{{n - 10 = 0}}} and/or {{{n + 12 = 0}}}
If {{{n - 10 = 0}}} then {{{n = 10}}} This is the first positive even integer.
If {{{n + 12 = 0}}} the {{{n = -12}}} Discard this solution as it is a negative integer.

The two integere are: 
n = 10  and n+2 = 12

10 and 12

Check:

{{{1/10 - 1/12 = 6/60 - 5/60}}} = {{{(6-5)/60 = 1/60}}}