If A is reducible to B and B is reducible to A, and A is NPC, is B also NPC?
First, note that there are two problems frequently called the "Maximum Clique Problem".

One of them is the Clique Decision Problem: given an undirected graph $G$ and a number $k$, does $G$ contain a clique with at least $k$ vertices? This is one of Karp's 21 Problems, the original list that made NP-completeness famous. The proof involved reducing to it from the Boolean Satisfiability Problem, the very first original NP-complete problem. And a verifier for this is easy to make: check whether the set given as a solution actually does contain $k$ or more vertices, and is actually a clique.

The other is the Maximum Clique Size Problem: given an undirected graph $G$, what is the size of the largest clique in $G$? This can be shown to be NP-hard by reduction from the Clique Decision Problem, by simply comparing the size of the largest clique with $k$. And as you mentioned, it can apparently be shown to be in NP by reduction to the Clique Decision Problem: for every integer from the number of vertices down to 1, check if there's a clique of that size, and if so return that number.

So it would seem that both versions of the problem are NP-complete. But be careful about the definition of NP. NP is defined as a class of decision problems, and the Maximum Clique Size Problem is not a decision problem.

Specifically, NP is the class of decision problems where, if the answer is yes, you can be given extra information of polynomial size that lets you verify that the answer is yes in polynomial time. (If the answer is no, you can't necessarily verify it, unless NP = co-NP.) So for problems which aren't answered with yes or no, trying to put them in NP just doesn't make sense.

So while the Maximum Clique Size Problem is indeed NP-hard, and can be reduced to an NP-complete problem in polynomial time, it is not in fact in NP. And thus it cannot be NP-complete. This is why you can't make a verifier for it (unless NP = co-NP, which is an open question).