Find a where 2(1/a+1/(a+1)) is a natural number.
It should be obvious that double the sum of two small fractions will be less than 1. In fact
$1 = 2 \left ( \dfrac{1}{4} + \dfrac{1}{4} \right ) \implies 1 > 2 \left ( \dfrac{1}{4 + k} + \dfrac{1}{5 + k} \right ) \text { if } k \ge 0.$
So you have exactly 3 numbers to test, 1, 2, and 3. And 2 and 3 don't work.