110 110 110 110 110 110 110 110 1
  101 101 101 101 101 101 101 101 1

The sum of N whole numbers being odd or even only depends on each of the numbers being odd or even. So we use 0 and 1, but:

• any 0 can be replaced with any even number
• any 1 can be replace with any odd number.

Since we need any 3 successive numbers to be even, we need a repeating pattern of 3 numbers.
The possible values of a 3 digit-pattern using only 0 and 1 are:
000 even
001 odd
010 odd
011 even
100 odd
101 even
110 even
111 odd

That leaves us with 000, 011, 101 and 110:
000 000 -> the sum of any 3 consecutive digits (always 000) is even.
011 011 -> the 3 consecutive digits can be either 011, 110 or 101, all 3 sums are even.
101 101 -> it's the same 3 groups just offset by 1
110 110 -> ditto

Another way of getting this list is considering that to get an even sum of 3 digits, we need either

:
 

Only even digits (0 -> 000)

Two odd digits (1, the sum of which will be even) and one even digit (0) -> 011, 101 or 110.

Now, by repeating 8 times either of these 4 patterns, we get a series of 24 numbers where the sum of any 3 consecutive numbers is even. Clearly the sum of all 24 numbers is even as well.

We still need to add one digit, and for the sum of all numbers to be odd. This condition means the last digit must be 1.
Since we need to continue repeating our pattern, it means the pattern must start with 1, which only leaves us 101 and 110.

110 110 110 110 110 110 110 110 1 
101 101 101 101 101 101 101 101 1

The sum of N whole numbers being odd or even only depends on each of the numbers being odd or even. So we use 0 and 1, but:

any 0 can be replaced with any even number

any 1 can be replace with any odd number.

Since we need any 3 successive numbers to be even, we need a repeating pattern of 3 numbers.
The possible values of a 3 digit-pattern using only 0 and 1 are:
000 even
001 odd
010 odd
011 even
100 odd
101 even
110 even
111 odd

That leaves us with 000, 011, 101 and 110:
000 000 -> the sum of any 3 consecutive digits (always 000) is even.
011 011 -> the 3 consecutive digits can be either 011, 110 or 101, all 3 sums are even.
101 101 -> it's the same 3 groups just offset by 1
110 110 -> ditto

Another way of getting this list is considering that to get an even sum of 3 digits, we need either:

Only even digits (0 -> 000)

Two odd digits (1, the sum of which will be even) and one even digit (0) -> 011, 101 or 110.

Now, by repeating 8 times either of these 4 patterns, we get a series of 24 numbers where the sum of any 3 consecutive numbers is even. Clearly the sum of all 24 numbers is even as well.

We still need to add one digit, and for the sum of all numbers to be odd. This condition means the last digit must be 1.
Since we need to continue repeating our pattern, it means the pattern must start with 1, which only leaves us 101 and 110.