Consider N = n_1 n_2 identical cells, each having an emf E and an...

Equivalent emf of one row of $n_1$ cells in series is $n_1 \mathcal{E}$ and its internal resistance is $n_1 r$.

Since $n_2$ such rows are connected in parallel, the equivalent emf of the entire combination remains $E_{eq} = n_1 \mathcal{E}$ and the equivalent internal resistance is $r_{eq} = \dfrac{n_1 r}{n_2}$.

The current $I$ through the external resistance $R$ is given by:

$I = \dfrac{E_{eq}}{R + r_{eq}} = \dfrac{n_1 \mathcal{E}}{R + \dfrac{n_1 r}{n_2}} = \dfrac{n_1 n_2 \mathcal{E}}{n_2 R + n_1 r}$

Since the total number of cells $N = n_1 n_2$ is constant, the numerator $N \mathcal{E}$ is constant. To maximize the current $I$, the denominator $n_2 R + n_1 r$ must be minimized.

We can rewrite the denominator as:

$n_2 R + n_1 r = \left( \sqrt{n_2 R} - \sqrt{n_1 r} \right)^2 + 2 \sqrt{n_1 n_2 R r}$

For the denominator to be minimum, the squared term must be zero:

$\sqrt{n_2 R} - \sqrt{n_1 r} = 0$

$n_2 R = n_1 r$

This gives the condition $r n_1 = R n_2$.