Answer
If $R=r$, then $i = \frac{N~\mathscr{E}}{(N+1)~r}$
In the circuit with the batteries in series and the circuit with the batteries in parallel, both arrangements give the same current in $R$ if $R = r$
Work Step by Step
We can consider the circuit with the batteries in series. We can find an expression for $i$, the current in $R$:
$\Delta V = N~\mathscr{E}-N~i~r-i~R = 0$
$N~\mathscr{E} = N~i~r+i~R$
$i = \frac{N~\mathscr{E}}{N~r+R}$
If $R=r$, then $i = \frac{N~\mathscr{E}}{N~r+r} = \frac{N~\mathscr{E}}{(N+1)~r}$
We can consider the circuit with the batteries in parallel. We can find an expression for $i$, the current in $R$. Note that the current across each battery is $\frac{i}{N}$. We can consider a single loop of the circuit:
$\Delta V = \mathscr{E}-\frac{i}{N}~r-i~R = 0$
$\mathscr{E} = \frac{i}{N}~r+i~R$
$i = \frac{\mathscr{E}}{\frac{r}{N}+R}$
$i = \frac{N~\mathscr{E}}{r+N~R}$
If $R=r$, then $i = \frac{N~\mathscr{E}}{r+N~r} = \frac{N~\mathscr{E}}{(N+1)~r}$
We can see that both arrangements give the same current in $R$ if $R = r$
