Proving NP completeness of maximal length path
I think here is a pitfall due to inaccurate notation. 
Notation for me:


A trail is a sequence of distinct connected edges.
A path is a trail with distinct vertices.


In the problem instance $MNPL$ (maximum neighbor path length) the sets $(G,N^+, N^-)$ and the weight function $c_e$ are given as input(and hereby fixed).
Since $N$ is partitioned, for two vertices $u,v$ either $\{u,v\}\notin E$ or $u$ is a positive neighbor of $v$ or vice-versa. Especially $u$ and $v$ cannot be a positive neighbor of each other by definition ($N$ would not be partitioned). Hence $G$ must be a directed graph. 
Now, for any path  $s \rightsquigarrow (u,v,w) \rightsquigarrow t$(not way nor cycle nor walk!) the neighbor-condition is fullfilled.

Maybe this was the real challenge of this exercise? 

So, claim: $MNPL$ is $NP$-complete.

Proof: $MNPL \leq_p LPP$ (longest path problem) 
Consider the reduction $f$. For $(G,N^+,N^-,c_e)$ construct $G'=(V',E',c_e')$ as the following:


$c_e'$ = $c_e$
$V' = V$
$E' = \bigcup_{v \in N^+(u), u \in N} (u,v) \cup \bigcup_{u \in N^-(v), v \in N} (v,u)$


Clear: $f$ is computable and runs in poly-time ($O(|E|)$). By construction and with the observations above $f$ will construct a digraph from $N$ and the neighbor sets.

So $f$ is a reduction and $MNPL \in NP$ is obvious using "guess and check".
Since $LPP$ is strongly $NP$-complete, $MNPL$ is as well strongly $NP$-complete.
This is no decision problem but a maximization problem.