Solution 1 (Generating Function, MarkBcc168): Consider the quantity $$T = (x+x^2+x^3+...)+(y+y^2+y^3+...) = \frac{x}{1-x}+\frac{y}{1-y}$$and define generating functions $$F(x,y) = 1+T+T^2+...$$It's clear that the coefficient of $x^ay^b$ in $F$ equals to the number of ways to jump with a total distance of $a+b$ and arrive at position $a-b$. (i.e. variable $x$ corresponds to positive jumps and variable $y$ corresponds to negative jumps). Now we evaluate $F(x,y)$ in $\pmod 2$. To do this, let $G(x,y) = 1-T+T^2-T^3+...$ so that $G\equiv F\pmod 2$ and $$G(x,y) = \frac{1}{1+T} = \frac{1}{1+\frac{x}{1-x}+\frac{y}{1-y}} = \frac{(1-x)(1-y)}{1-xy}$$Thus, we have $$G(x,y) = (1-x-y+xy)(1+(xy)+(xy)^2+(xy)^3+...)$$It's clear that all odd coefficients are in form $x^ny^{n+1}$ and $x^{n+1}y^n$, which corresponds to $N(1)$ and $N(-1)$. Thus the answer is $\boxed{\{1,-1\}}$. Solution 2 (Combinatorial, Official): Encode each positive jump by the corresponding number of $\texttt{+}$ and encode each negative jump by the corresponding number of $\texttt{-}$. We also seperate each jump by $\texttt{|}$. For instance, $$\texttt{++|+++|---}\implies 0\stackrel{+2}{\longrightarrow} 2\stackrel{+3}{\longrightarrow} 5\stackrel{-3}{\longrightarrow} 2.$$Clearly, in $N(a)$, we must have $\frac{2019+a}{2}$ $\texttt{+}$'s and $\frac{2019-a}{2}$ $\texttt{-}$'s. We also note that a $\texttt{|}$ must be inserted between $\texttt{+-}$. Now, fix a sequence consisting many $\texttt{+}$ and $\texttt{-}$. Call a sequence \emph{bad} if and only if there are odd number of ways to insert $\texttt{|}$. Claim: The only bad sequences are $\texttt{+-+-+...+}$ and $\texttt{-+-+-...-}$. Proof: If $m,n$ denote the number of consecutive $\texttt{++}$ and $\texttt{--}$ respectively. Then clearly the number of ways to insert $\texttt{|}$ is precisely $2^{m+n}$. Thus the sequence is bad if and only if there are no $\texttt{++}$ and $\texttt{--}$ at all so we are done. The two bad sequences correspond to $N(1)$ and $N(-1)$ thus the answer is $\{1,-1\}$.