The smallest possible sum (a+b)= 36+45=81
1/20= 1/a+1/b= (a+b)/ab --> 20(a+b)=ab
b=20a/(a-20)=20+400/(a-20)
Hence a+b=a+20-400/(a-20),
By differentiation, a+b is minimized when (a-20)^2=400, i.e. a=40, b=40.
However, this does not fit the requirement a<b. The solution pair that is closest to this pair would be the solution.
Consider b=20+400/(a-20) and b is an integer,
(a-20) has to be a factor of 400. As 16 is a factor of 400 and none of 17,18,19 is factor of 400, we can see that (a,b)=(36,45) is the pair of solution we are looking for.