If we have a sequence from 0 to N in binary format then the number of 0 and number of 1s in the least significant bit have a balance i.e. if N was even then there is one 0 more than how many 1s are there. If it is odd the number of 0s and 1s are the same.
Example:
0000
0001
0010
0011
There are 2 0s and 2 1s in the first binary column (N = 3 i.e. odd)
0000
0001
0010
0011
0100
There are 3 0s and 2 1s in the first binary column (N = 4 i.e. even).
This makes sense since each number is basically the previous +1, so each addition either adds a 1 or a 0 to the last column and since this is about all the numbers from 0 to N this equation is there.
What is not straightforward to me to understand intuitively is that if I split these numbers to 2 sets, i.e. odds (or those with the last bit set to 1) and evens (those with the last bit set to 0) the same equation holds.
The same also holds for all further subdivisions of those with the second column 0 vs second column 1 etc.
Could some give some insight why this further and further subdivision keeps having this condition hold?
The rightmost two bits go through the cycle $00, 01, 10, 11$, which corresponds to counting $0,1,2,3$. If you separate out the even numbers, you select those with $0$ in the ones bit, and they will count $0,2,0,2,0,2$. The twos bit will alternate between $0$ and $1$, just like the ones bit did when you had all the numbers. It works the same however many bits you select and which selection you make. If you just take the numbers that end in $101$ you will get the numbers that are equivalent to $5 \bmod 8$. If you consider the next bit, which is the eights place, you will alternate between $5 \bmod 16$ and $13 \bmod 16$. The first has $0$ in the eights place and the second has $1$ in the eights place, so you get the same alternation.