We have a 6 digit number _ _ _ _ _ _
the constraints are :
1. the 6th digit is the sum of 2th and the 4th digit.
2. the 5th digit is the sum of 1th and the 3th digit.
3. digits can be from 1 to 9. Repetition of digits is allowed.
how many possibilities are there?
I have tried something, written down every possibly combination of sum of two digits that doesn't exceed 9. I got 36. then multiplied it by 4. but it's not correct
Let's call the digits $d_1, d_2, \ldots d_6$. Clearly the first four digits determine the whole number, but there are constraints on those digits in order to ensure that $d_5$ and $d_6$ stay in range.
So, for example, choosing $d_1=1$ gives $8$ options for $d_3$ to keep $d_5$ in range. Choosing $d_1=2$ gives $7$ options for $d_3$, etc., leading to a total of $8+7+6+\cdots+1=36$ options for the $d_1,d_3$ set. This agrees with part of your reasoning.
Independently of that choice, by the same calculations we similarly have $36$ options for $d_2,d_4$. Clearly any one choice for $d_1,d_3$ can be associated with any choice for $d_2,d_4,$ so overall the total choices for the whole six digit number are $36\cdot 36 = 1296$.