We consider OP's problem in a slightly more convenient (symmetrical) setting:
Let $0\leq n\leq m$. We are looking for the number $L_{m,n;r,s}$ of lattice paths starting in $(0,0)$ and ending in $(m,n)$ which do not reach the lines $y=r$ and $y=-s$ where $r,s>0$. Admissible steps are $(1,1)$ and $(1,-1)$.
We show the following is valid:
\begin{align*}
\color{blue}{L_{m,n;r,s}}&\color{blue}{=\binom{m}{\frac{m+n}{2}}-\sum_{j\geq0}\left[\binom{m}{\frac{m+n}{2}-(j+1)r-js}
+\binom{m}{\frac{m+n}{2}+jr+(j+1)s}\right]}\\
&\qquad\qquad\qquad\color{blue}{+\sum_{j\geq1}\left[\binom{m}{\frac{m+n}{2}+j(r+s)}
+\binom{m}{\frac{m+n}{2}-j(r+s)}\right]}\tag{1}\\
\end{align*}
Note, the sums in (1) are finite since $\binom{p}{q}=0$ if $q<0$ or $q>p$. OPs problem is looking for the number of paths from $(0,0)$ to $(N-1,r-1)$ which do not reach the lines $y=r$ and $y=-(l+1)$, so that (1) can be applied with
\begin{align*}
m&=N-1\\
n&=r-1\\
s&=l+1\\
\end{align*}
We prove (1) in three steps. At first we are looking for the number of paths from $(0,0)$ to $(m,n)$ without boundary restrictions using steps $(1,1)$ and $(1,-1)$.
Step 1: The number $L_{m,n}$ of paths from $(0,0)$ to $(m,n)$ where $0\leq n\leq m$ is
\begin{align*}
\color{blue}{L_{m,n}=\binom{m}{\frac{m+n}{2}}}\tag{2}
\end{align*}
We show (2) algebraically. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We encode the steps $(1,1)$ with $xy$ and $(1,-1)$ with $\frac{x}{y}$. We obtain
\begin{align*}
L_{m,n}&=[x^my^n]\left(xy+\frac{x}{y}\right)^m\tag{3}\\
&=[x^my^n]x^my^{-m}\left(1+y^2\right)^m\\
&=[y^{m+n}]\sum_{j=0}^m\binom{m}{j}y^{2j}\tag{4}\\
&=\binom{m}{\frac{m+n}{2}}\tag{5}
\end{align*}
and the claim follows.
Comment:
In (3) we note that each step is either $(1,1)$ or $(1,-1)$ which can be encoded as $xy+\frac{x}{y}$.
In (4) we expand the binomial and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.
In (5) we select the coefficient of $y^{m+n}$. We also note according to the specific steps $(1,1)$ and $(1,-1)$ the parity of $m$ and $n$ is the same so that $\frac{m+n}{2}$ is always an integer.
Step 2: The number $L_{m,n;r,-}$ of paths from $(0,0)$ to $(m,n)$ where $0\leq n\leq m$ which do not reach the boundary $y=r$ with $r>0$ is
\begin{align*}
\color{blue}{L_{m,n;r,-}=\binom{m}{\frac{m+n}{2}}-\binom{m}{\frac{m+n}{2}-r}}\tag{6}
\end{align*}
We prove (6) using Andre's reflection principle. The number of all paths from $(0,0)$ to $(m,n)$ is $L_{m,n}$. We subtract all invalid paths which are those reaching the line $y=r$. An invalid path touches (or crosses) the line a first time. We reflect each invalid path part from the origin to the first contact with $y=r$ at $y=r$ and obtain all paths from $(0,2r)$ to $(m,n)$.
Denoting with $L[(0,2r),(m,n)]$ the number of all invalid paths we have
\begin{align*}
L[(0,2r),(m,n)]&=L_{m,n-2r}=\binom{m}{\frac{m+n}{2}-r}
\end{align*}
and the claim (6) follows.
Step 3: The number $L_{m,n;r,s}$ with boundaries at $y=r$ and $y=-s$ is given by (1).
This number can be calculated using Andre's reflection principle in conjunction with the inclusion-exclusion principle (PIE).
We denote with $L(A_1)$ the paths which reach $y=r$, with $L(A_2)$ the paths which reach $y=r$, then $y=-s$ in that order, with $L(A_3)$ the paths which reach $y=r$, then $y=-s$, then $y=r$ in that order, etc.
Analogously w denote with $L(B_1)$ the paths which reach $y=-s$, with $L(B_2)$ the paths which reach $y=-s$, then $y=r$ in that order, with $L(B_3)$ the paths which reach $y=-s$, then $y=r$, then $y=-s$ in that order, etc.
Application of PIE as compensation for double counting gives
\begin{align*}
\color{blue}{L_{m,n;r,s}=\binom{m}{\frac{m+n}{2}} + \sum_{j\geq 1}(-1)^j\left(L(A_j)+L(B_j)\right)}\tag{7}
\end{align*}
We find by application of the reflection principle
\begin{align*}
L(A_1)&=L\left[(0,2r),(m,n)\right]=L_{m,n-2r}=\binom{m}{\frac{m+n}{2}-r}\\
\color{blue}{L(A_{2j+1})}&=L\left[(0,2(j+1)r+2js),(m,n)\right]=L_{m,n-2(j+1)r-2js}\\
&\,\,\color{blue}{=\binom{m}{\frac{m+n}{2}-(j+1)r-js}}\qquad\qquad \color{blue}{j\geq 0}\tag{8}\\
L(A_2)&=L\left[(0,-2r-2s),(m,n)\right]=L_{m,n+2r+2s}=\binom{m}{\frac{m+n}{2}+r+s}\\
\color{blue}{L(A_{2j})}&=L\left[(0,-2jr-2js),(m,n)\right]=L_{(m,n+2jr+2js}\\
&\,\,\color{blue}{=\binom{m}{\frac{m+n}{2}+jr+js}}\qquad\qquad\qquad\ \ \color{blue}{j\geq 1}\tag{9}\\
\end{align*}
Analogously we obtain
\begin{align*}
L(B_1)&=L\left[(0,-2s),(m,n)\right]=L_{m,n+2s}=\binom{m}{\frac{m+n}{2}+s}\\
\color{blue}{L(B_{2j+1})}&=L\left[(0,-2jr-2(j+1)s),(m,n)\right]=L_{m,n+2jr+2(j+1)s}\\
&\,\,\color{blue}{=\binom{m}{\frac{m+n}{2}+jr+(j+1)s}}\qquad\qquad \color{blue}{j\geq 0}\tag{10}\\
L(B_2)&=L\left[(0,+2r+2s),(m,n)\right]=L_{m,n-2r-2s}=\binom{m}{\frac{m+n}{2}-r-s}\\
\color{blue}{L(B_{2j})}&=L\left[(0,2jr+2js)\right]=L_{m,n-2jr-2js}\\
&\,\,\color{blue}{=\binom{m}{\frac{m+n}{2}-jr-js}}\qquad\qquad\qquad\ \ \color{blue}{j\geq 1}\tag{11}\\
\end{align*}
Finally putting (7) - (11) together we get the claim (1).
Example:
Now it's time to harvest. So, let's make an example which can also be easily manually checked. We look for the number of paths from $(0,0)$ to $(14,2)$ which do not reach the boundary lines $y=4$ and $y=-3$.
This number is $\color{blue}{L_{14,2;4,3}=1\,652}$ which is marked red in the graphic below.
Applying (1) we obtain
\begin{align*}
\color{blue}{L_{14,2;4,3}}&=\binom{14}{8}-\sum_{j\geq0}\left[\binom{14}{8-4(j+1)-3j}
+\binom{14}{8+4j+3(j+1)}\right]\\
&\qquad\qquad\qquad+\sum_{j\geq1}\left[\binom{14}{8+7j}
+\binom{14}{8-7j}\right]\\
&=\binom{14}{8}-\left[\binom{14}{4}+\binom{14}{11}\right]+\left[\binom{14}{1}\right]\tag{12}\\
&=3\,003-\left(1\,001+364\right)+\left(14\right)\\
&\,\,\color{blue}{=1\,652}
\end{align*}
in accordance with the manual calculation in the graphic.
In (12) we have two summands in brackets. They give the number of reflected paths indicated in the graphic via $A_1$ and $B_1$.
The right-most summand in (12) gives the number of reflected paths indicated in the graphic via $B_2$.
No more reflections need to be considered in this example.
