Proving a problem is NP Hard
Let $G$ be a graph with $2n$ vertices, and choose any node $v$. The edges touching $v$ have weight $n$, and all other edges have weight $-1$. Note that any Hamiltonian cycle in $G$ will have two edges touching $v$ (which add up to $2n$) and $2n-2$ edges of weight $-1$, so the total weight of the cycle is exactly $2$.
Therefore, if $G$ has a hamiltonian cycle, it has a cycle of weight exactly $2$.
Moreover, assume $G$ has a cycle of weight exactly $2$. Such a cycle can either include $v$ or not. If it doesn't include $v$, all its edges will be negative, contradicting a total weight of $2$. If it includes $v$, then it must include two edges of weight $n$, and thus the remaining part of the cycle must be a path, using only edges of weight $-1$, and with total weight equal to $-2n+2$. That meeans such a path must contain $2n-2$ edges, and therefore have $2n-1$ nodes. When connecting said path to-and-from $v$, you have a Hamiltonian cycle.
This means that if you could decide in polynomial time whether $G$ has a cycle of weight $2$ (which is a fixed constant), you could decide whether it has a Hamiltonian cycle in polynomial time.