To maximize the sum, start by putting 9, 8 and 7 in the corners so they can each be used on two sides. That leaves sides of sums 15, 16, and 17, each with two blanks to fill in. The sum of the remaining numbers is 21, and we need to split them into three pairs with sums that are three consecutive numbers.

21 as the sum of three consecutive numbers is 6+7+8. There are multiple ways of splitting the numbers 1-6 into three pairs whose sums are 6, 7, and 8. One such way is 2+4, 1+6, and 3+5. The resulting triangle is

9 2 4 8
 6   5
  1 3
   7

Basically, match up the pair totaling 6 with the pair totaling 17. Do the same for 7 with 16, and 8 with 15. The maximum possible sum on each side of the triangle is 23.