If $n_l(t)$ is the number of particles in the left half after $t$ moves, the change in $n_l(t)$ in the time interval $\Delta t$ is $$\begin{equation*} \Delta n=\left(\frac{N-n_l(t)}{N}-\frac{n_l(t)}{N}\right)\Delta t, \end{equation*}$$ and assuming that $n_l$ and $t$ are continuous variables we arrive at $$\begin{equation*} \frac{dn_l(t)}{dt}=1-\frac{2n_l(t)}{N}, \end{equation*}$$ whose solution is $$\begin{equation*} n_l(t)=\frac{N}{2}\left(1+e^{-2t/N}\right), \end{equation*}$$ with the initial condition $n_l(t=0)=N$. Note that we have assumed $n$ to be a continuous variable. Obviously, particles are discrete objects.