So far, all the answers have taken the following form: you just biject N with some countably infinite group. But what if one regarded that as "cheating" and asked for a truly "natural" group operation on N? So far, I think Pietro Majer's answer comes closest, since his bijection with an infinite product of cyclic groups of order 10 is fairly natural (at least in the sense of "familiar"). But can one do any better than this?
Here is an attempt to argue that there is no 100% natural group structure on N. (I don't claim that this is surprising.)
First, we need to choose an identity element. The identity element is a privileged element of the group, so for the choice to be natural it should be a privileged element of N. The only privileged element of N is 1 (for me the natural numbers start at 1) so we are forced to choose that. Now let's consider the following question: what is the inverse of 2? I think there are two possible natural answers, namely 2 itself and 3. (One is the smallest element you can choose, and the other is the smallest element that is not equal to 2.) If we choose 2 and continue, then perhaps we'll end up with the binary version of PM's answer (except that everything would be shifted by 1 because of my insistence on making 1 the smallest natural number -- perhaps that was silly). So perhaps, contrary to my original intention, that is an argument that PM's answer is (apart from taking base 10) the most natural thing one can do, and not too unnatural.
If we go for 3, then there doesn't seem to be any natural choice for the order of 2 other than infinity. So then we need to keep powers of 2 away from powers of 3. The natural way of doing that looks like actually using the multiplicative structure of N. But then we get into trouble later, since 2*3 will be 1, so our treatment of 6 becomes a bit arbitrary. Or at least, I think it does. An implicit question here: is there some optimally natural way of putting a group structure on N in such a way that every element has infinite order? Perhaps the bijection-with-Z answer is the best one can do.