Each digit must appear exactly once. The two boxes that total 4 must be 1 and 3. One of the two boxes that total 12 is either a 1 or a 3; 9 is the highest number we can use, so the 12 must be 3 + 9. Put 3 in the lower right hand corner, put 1 to the left of the 3, and put 9 above the 3.

The four boxes that add up to 20 and contain the 9 (upper right) must include the three lowest numbers remaining (2, 4, and 5); any higher numbers would total to more than 20. That means the left column contains 6, 7, and 8 (the only other numbers).

Place the digits 1 – 9 in the nine boxes below, one digit per box. The numbers in the circles indicate the total of the digits in each of the boxes that the circle touches.

The four boxes in the lower left corner add up to 16. The lowest possible numbers are the only possible choices that total 16. Therefore, 2 is in the center square, and 6 and 7 are in the other two squares. That leaves 8 in the upper left corner.

The four squares in the upper left corner add up to 20. The 2 and the 8 are already known; the lowest numbers remaining are 4 and 6. Any other choices total more than 20. Therefore, the final answer is 8, 4, and 5 on the top row, 6, 2, and 9 on the middle row, and 7, 1, and 3 on the bottom row.