Here, let me try to redeem myself from my first post. First, let me say there is some ambiguity in the question. Clearly the number 111 is a palindrome for base 2 - infinity. For base 2, the number is equal to 7, base 3 it is equal to 13, base 4 it is equal to 21, etc. I am going to assume that is not what the question is asking, because that is just silly.
I assume the question asks is there a number that can be converted into 2002 palindromes?
I believe the answer is yes.
I recognize that 0x0 is a convenient palindrome. As an example, take 100, which can be represented as 010 in base 100, 020 in base 50, 040 in base 25, and 050 in base 20. There are other palindromes in different formats (e.g., 121 in base 9 is still 100) but I’m not going to worry about those.
The larger the number gets, the more 0x0 palindromes you have. For example, 1000 is 0 1 0 in b(1000), 0 2 0 in b(500), 0 4 0 in b(250), 0 5 0 in b(200), 0 8 0 in b(125), 0 10 0 in b(100), 0 20 0 in b(50), and 0 25 0 in b(40).
Note that in both cases the base b can not be greater than or equal to the square root of the integer in question. It does not make sense to have a number 0 10 0 represent 100 in base 10.
The general statement can be made that a number N has at least Z number of palindromes, where Z equals the number of unique factors that are less than the sqrt(N).
Therefore, there clearly exists a finite number less than infinity that has 2002 unique factors in it whereby each of those factors are less than the square root of N.