Solution:

When $N$ identical cells, each of emf $e$ and internal resistance $r$ , are connected in series and $n$

cells are connected in reverse, the resulting emf and internal resistance of the battery can be determined as follows:

1. Resultant emf ( $\varepsilon_0$):

• For correctly connected cells, the total emf is:
  $$\varepsilon_{\text{correct}} = (N - n)e.$$ 
• For n reversed cells, their emf opposes the total emf. The opposing emf is:$$\varepsilon_{\text{reverse}} = ne.$$The net emf of the resulting battery is:

  $$\varepsilon_0 = \varepsilon_{\text{correct}} - \varepsilon_{\text{reverse}} = (N - n)e - ne = (N - 2n)e.$$ 

2. Resultant internal resistance ( $r_0$):

• All cells, whether correctly or incorrectly connected, contribute to the total internal resistance because resistances add in series. The total internal resistance is:
  $$r_0 = N r.$$ 

Final Answer:

The emf and internal resistance of the resulting battery are:

$$\boxed{\varepsilon_0 = (N - 2n)e, \quad r_0 = Nr.}$$