Given array with n numbers, it is known that 9 appears at-least n-n^(99/100) times in the array. Is there comparison-sort algorithm in O(n) time?

Question

Problem: We are given an array with n numbers and it is known that the number 9 appears at-least n-n^(99/100) times in the array.
Prove/Disprove: There exists a comparison-based algorithm that sorts the given array in linear time.

Note: comparison-sort based algorithm is discussed and not non-comparison-sort algorithms like counting sort, radix sort...

Attempt: I think the answer is a disproof but I don't really know how to justify my claim. I learned that the lower-bound at worst-case on comparison-sort algorithms is Ω(nlogn) , thus a lower bound for sorting rest of the n^(99/100) elements in the above array at worst case will be Ω( n^(99/100)log(n^(99/100))) = Ω( n^(99/100)log(n)) which is not linear, after sorting we place the 9 in the appropriate index, that'll cost O(n^(99/100)) at worst-case ( we're going over all the elements in the up-to now sorted array in order to find the neccessary index ).
( I don't know what other important information I can deduce from the problem, so I'm stuck )

Thanks in advance for help!