Issue about the generating function 
A rabbit initially stands at the position $0$, and repeatedly
jumps on the real line. In each jump, the rabbit can jump to any
position corresponds to an integer but it cannot stand still. Let
$N(a)$ be the number of ways to jump with a total distance of
$2019$ and stop at the position $a$. Determine all integers $a$
such that $N(a)$ is odd.

Solution 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 $x^my^n$ in $F$ equals to the number of ways to
jump with a total distance of $m+n$ and arrive at position $m-n$. (i.e. variable $x$ corresponds
to positive jumps and variable $y$ corresponds to negative jumps).
Now we evaluate $F(x,y)$. To do this, we work in $\mathbb{Z}_2$, so
$$F(x,y) = \frac{1}{1-T} = \frac{(1-x)(1-y)}{1-xy}$$
Thus, we have
$$F(x,y) = (1-x-y+xy)(1+(xy)+(xy)^2+(xy)^3+...)$$ 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 $\boxed{\{1,-1\}}$.
Edit after Donald Splutterwit answer. Can someone please explain


*

*What is $1$ in $F$ since rabbit must jump at least once? >Shouldn't it start with $T$ and not with $1$?

*Why there is no $1$ in $T$ since the rabbit can jump on the >spot?

*How come they never actually use $2019$ and $a$?




Question: What is $1$ in $F$ since rabbit must jump at least once? Shouldn't it start with $T$ and not with $1$?

We are free to choose $F$ either starting with $1$ or starting with $T$. A value $1$ represents no jump at all and will simply be ignored when doing the analysis. Its benefit is to have a slightly simpler function. With
\begin{align*}
T(x,y)=\frac{x}{1-x}+\frac{y}{1-y}
\end{align*}
we have
\begin{align*}
\color{blue}{F(x,y)}&=\frac{1}{1-T(x,y)}=\frac{1}{1-\left(\frac{x}{1-x}+\frac{y}{1-y}\right)}\\
&=\frac{1-x-y+xy}{1-\left(2x+2y-3xy\right)}\\
&=\left(1-x-y+xy\right)\sum_{j=0}^\infty \left(2x+2y-3xy\right)^j\\
&\color{blue}{\equiv\left(1+x+y+xy\right)\sum_{j=0}^\infty\left(xy\right)^j\pmod{2}}\tag{1}\\
\end{align*}
On the other hand starting with $T$ we obtain
\begin{align*}
\color{blue}{G(x,y)}&=\frac{T(x,y)}{1-T(x,y)}=\frac{\frac{x}{1-x}+\frac{y}{1-y}}{1-\left(\frac{x}{1-x}+\frac{y}{1-y}\right)}\\
&=\frac{x+y-2xy}{1-\left(2x+2y-3xy\right)}\\
&=\left(x+y-2xy\right)\sum_{j=0}^\infty \left(2x+2y-3xy\right)^j\\
&\color{blue}{\equiv\left(x+y\right)\sum_{j=0}^\infty\left(xy\right)^j\pmod{2}}\tag{2}\\
\end{align*}
When answering the third question, we will see that (1) and (2) will lead to the same conclusion.

Question: Why there is no $1$ in $T$ since the rabbit can jump on the >spot?

The problem states the rabbit can jump to any position corresponding to an integer but it cannot stand still. This implies that each jump has length $>0$.

Question: How come they never actually use $2019$ and $a$?

The solution answers the prolem for any distance $d$, not only for the special case $d=2019$. We consider wlog the case $a>0$ which implies when looking at positive jumps $x^m$ and negative jumps $y^n$ we have
\begin{align*}
m+n&=d\tag{3}\\
m-n&=a
\end{align*}
We denote with $N(d,a)$ the number of ways to jump with a total distance of $d$ and stop at the position $a$.
We use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ in a series. We obtain calculating in $\mathbb{Z}_2$
\begin{align*}
\color{blue}{[x^my^n]}&\color{blue}{F(x,y)}\\
&=[x^my^n]\left(1+x+y+xy\right)\sum_{j=0}^{\infty}(xy)^j\\
&=\left([x^my^n]+[x^{m-1}y^n]+[x^my^{n-1}]+[x^{m-1}y^{m-1}]\right)\sum_{j=0}^\infty (xy)^j\\
&\color{blue}{=\begin{cases}
1&\qquad (m=n)\vee (m=n+1)\vee (m=n-1)\\
0&\qquad\text{otherwise}
\end{cases}}\tag{4}
\end{align*}

An odd coefficient $N(d,a)$ is given if
\begin{align*}
[x^my^n]F(x,y)=1\qquad\qquad m+n=d, m-n=a
\end{align*}
We conclude from (4) that $m=n$ implies $a=0$ which is not admissible. The other left options are $m-n=a=1$ and $m-n=a=-1$, resulting in
\begin{align*}
\color{blue}{N(d,-1),N(d,1)}
\end{align*}
for any odd values $d$. No other value $a$ is possible.

This solution implicitly includes the value $d=2019$.
We also note that working with $G(x,y)$ results in (4) without the solution $m=n$ which is not admissible anyways.
