Amortize time for a counter with the operations INCREMENT and DECREMENT
Your solution is on track. As you comment, if you increment a counter $2^k-1$ times and then do $m$ increment/decrement operations, in total you must have modified at least $km$ array positions (this will serve as our lower bound on the running time). In your case $N = 2^k-1 + m$. If you choose, for example, $m = 2^k+1$, then $N = 2^{k+1}$ while the lower bound is $km \geq \log_2 N \cdot (N/2) = \Omega(N\log N)$. This shows that the amortized running time is $\Omega(\log N)$, which is certainly superconstant.

We can easily get a matching (non-amortized) upper bound. If we only perform $N$ operations, then the value of the counter is in the range $[-N,N]$, and so takes $O(\log N)$ bits to represent. It is not difficult to implement increment and decrement so that they take time linear in the length of the representation, so $O(\log N)$ in this case.