Let the age of one surfer be 10a+b, then the age of the other is 10b+a. Add them together: 11a+11b, so 1/11 is a+b=sqrt(10b+a-10a-b)+1=sqrt(9b-9a)+1=sqrt(9(b-a))+1=3sqrt(b-a)+1. If b-a is to be a perfect square, the only ones are 0, 1, 4, 9, making a+b=1, 4, 7, 10. Now there's a problem: if we use the elimination method using two equations b+a=M and b-a=N, where M and N are positive integers, we add the equations together to eliminate a and we get 2b=M+N, so b=(M+N)/2. Therefore M and N must be both odd or both even. Unfortunately, if M is in the set 1, 4, 7, 10 then the corresponding N values are 0, 1, 4, 9. But when M is even N is odd or vice versa, and we would end up with fractional values for a and b.
If the question is interpreted as a+b=sqrt(9(b-a)+1) (the 1 is added before taking the square root instead of after), then the only perfect square occurs when b-a=7, so we have sqrt(9*7+1)=8=a+b. Again, an odd and an even number. (There is actually another trivial solution: when a=b, a+b=1, and we still end up with fractions!)
There is a solution if "plus 1" is omitted: 15 and 51 added together make 66; divide by 11 to get 6; subtract 15 from 51=36, the square root of which is 6.