Thanks for drawing attention to this corner case!

Right, so it counts as the full '2' (as opposed to being only 'worth' 1.

My code Java throws a divide-by-zero exception in the main test suite for the random big numbers. Careful with the overflow :)

787 is both palindrome and two before 789...

I'm sorry, but I can't comprehend the phrase "at least as long as when evenly divided".

The answer obtained by googling your suggested phrase goes to a stackoverflow question about doing this in python. The two most common approaches there split a list either as

10 10 10 10 10 10 10 5

which is simply the result of partition a 75-long list into chunks of 10 plus the leftover and has nothing to do with equal splitting,
or as

6 5 5 5 5

It's easy to see, that any list can be split into any amount of chunks which differ in length by no more than 1. Here's an example:

1 -> 1 0 0 0 0
2 -> 1 1 0 0 0
3 -> 1 1 1 0 0
4 -> 1 1 1 1 0
5 -> 1 1 1 1 1
6 -> 2 1 1 1 1

Is it a requirement that the first two chunks be equal in length? What about 1 -> 1 0 0 ...?

Because, honestly, I do not see how 6 -> 2 2 1 1 0 is more desirable. It seems to break the pattern, though it is much more like 11 -> 3 3 3 2 0 (but why not 3 3 2 2 1?).