1. Count total valence electrons in $\mathrm{NO_2^-}$

• Nitrogen has $5$ valence electrons.
• Each oxygen has $6$ valence electrons, so two oxygens contribute $2 \times 6 = 12$.
• The negative charge adds $1$ extra electron.

Thus, total valence electrons are $5 + 12 + 1 = 18.$

2. Draw the Lewis structure

The nitrite ion has nitrogen as the central atom: $\mathrm{O-N-O}$

To satisfy octets, one common Lewis structure is:

• one $\mathrm{N=O}$ double bond,
• one $\mathrm{N-O^-}$ single bond,
• one lone pair on nitrogen.

There are resonance structures, but in each Lewis structure nitrogen has:

• one double bond $\Rightarrow 4$ electrons around N,
• one single bond $\Rightarrow 2$ electrons around N,
• one lone pair $\Rightarrow 2$ electrons around N.

3. Count electrons around nitrogen

So total number of valence electrons around nitrogen is $4 + 2 + 2 = 8.$

Hence, nitrogen satisfies the octet.

Final integer answer: $\boxed{8}$