Step 1: Learn the two tests.
Isoelectronic means the two species carry the same total number of electrons. Isostructural means they take the same shape. We need a pair that passes both tests at once, so we count electrons first and then decide shape.
Step 2: Count electrons for the ammonium and borohydride pair.
For $\text{NH}_4^+$, add the electrons of N (7) and four H (4) and remove one for the positive charge, giving 10. For $\text{BH}_4^-$, add B (5) and four H (4) and add one for the negative charge, giving 10. Both have 10 electrons, so they are isoelectronic.
Step 3: Check their shapes.
Both $\text{NH}_4^+$ and $\text{BH}_4^-$ have a central atom bonded to four hydrogens with no lone pairs left over. Four bonding pairs and zero lone pairs always give a tetrahedral shape. So both are tetrahedral, meaning they are isostructural too.
Step 4: Knock out the ozone and nitronium pair.
$\text{O}_3$ is bent because the central oxygen has a lone pair, while $\text{NO}_2^+$ is linear with no lone pair on nitrogen. Same electrons maybe, but different shapes, so they fail the isostructural test.
Step 5: Knock out the remaining pairs.
$\text{N}_2\text{O}$ and $\text{NO}_2$ differ in both electron count and shape. $\text{NH}_2^-$ is bent (two lone pairs) while $\text{BH}_4^-$ is tetrahedral, so they cannot be isostructural. Both pairs are rejected.
Step 6: Lock the answer.
Only the first pair passes both checks, same 10 electrons and same tetrahedral shape.
\[ \boxed{\text{NH}_4^+ \text{ and } \text{BH}_4^- \text{ are isoelectronic and isostructural}} \]