A Louis Vuitton store in Shanghai had a number of pairs of sunglasses which cost an average of \$$900$ per pair. LiAngelo Ball stole a pair which cost \$$2000$. Afterwards, the average cost of sunglasses in the store dropped to \$$890$ per pair. How many pairs of sunglasses were in the store before LiAngelo Ball stole? 

[i]2018 CCA Math Bonanza Individual Round #3[/i]
1. Let \( n \) be the number of pairs of sunglasses in the store before the robbery.
2. The total cost of all the sunglasses before the robbery is \( 900n \) dollars.
3. After LiAngelo Ball stole a pair of sunglasses costing \( 2000 \) dollars, the total cost of the remaining sunglasses is \( 900n - 2000 \) dollars.
4. The number of pairs of sunglasses left in the store is \( n - 1 \).
5. The average cost of the remaining sunglasses is given as \( 890 \) dollars per pair.
6. Set up the equation for the average cost after the robbery:
   \[
   \frac{900n - 2000}{n - 1} = 890
   \]
7. Solve for \( n \):
   \[
   900n - 2000 = 890(n - 1)
   \]
8. Distribute \( 890 \) on the right-hand side:
   \[
   900n - 2000 = 890n - 890
   \]
9. Subtract \( 890n \) from both sides:
   \[
   900n - 890n - 2000 = -890
   \]
10. Simplify the equation:
    \[
    10n - 2000 = -890
    \]
11. Add \( 2000 \) to both sides:
    \[
    10n = 1110
    \]
12. Divide both sides by \( 10 \):
    \[
    n = 111
    \]

The final answer is \( \boxed{111} \).