What is the complexity class of arranging $1,\ldots, n$ so that all neighbors sum to a perfect square?
Whether this problem is in NP or not (or rather, in NP or not known to be in NP) depends on how you encode the input. If you encode the input in binary (or in decimal), then the input "$N$" has length $n = O(\log N)$, whereas the obvious witness has length $\Theta(N\log N)$, which is exponential in $n$ (of course, there could be a more compact witness), giving a complexity upper bound of only NEXP. In contrast, if you encode the input in unary (that is, as $1^N$, i.e., $N$ times the symbol $1$), then the input length is $N$ as well, and so the witness has polynomial length, putting the problem in NP.

Even if you use the second encoding, it seems unlikely that the problem is NP-complete. Most NP-complete problems involve a lot of data, and even the few problems which have a constant number of integers as input have more than one integer as input. Of course, formally, speaking, you can think of the input to any problem as a single integer. But semantically, there is a big difference.

What is the complexity of the problem, then? If there is a nice characterization of the values of $N$ for which such an arrangement exists, then it is likely that implementing the characterization would put the problem in P (even using binary encoding). As an extreme example, if such an arrangement exists for all large enough $N$, or doesn't exist for all large enough $N$, then the problem can be solved in linear time. If there is no simple characterization, then the problem might indeed be difficult, but probably not NP-complete or NEXP-complete.