Completed grid:
Logic:
Let's look at the bottom row first.
Consider the location of the 2 on the bottom row. It cannot be in entry 2 (same diagonal as 2 in the top row), position 3 or position 8 (same diagonal as a 1). If it were in position 5, then R1P5 would have to be 10, since the two diagonals on R2P5 have to have the same sum. [EDIT: The original logic (which follows) had an error in it...this actually shows that R1P4 is 6 greater than R3P3, not the other way around: but then the two diagonals on R2P4 would have to have the same sum, which implies R3P3 is 6 greater than R1P4, which is at least 3.] This implies that we must have R3P3 = 3 and R1P4 = 9: 1 and 2 are already used in the third row, and 10 is already used in the top row. But R2P2 must equal R3P3 to keep the diagonal sums on R1P2 equal, which forces R2P2 = 3, a contradiction given the 2 in its diagonal. So R3P7 must be 2.
Now let's look for the 3 in the bottom row. R3P2 cannot be 3, because it is on the same diagonal as the 2 in R1P2. From the red square in R1P3, we know that R2P2 and R3P3 have to be equal, so R3P3 cannot be 3 from the same diagonal. Note that R2P6 cannot be 7-9 from the 8 in R1P6, cannot be 5-6 from the 6 in R3P6, and cannot be 1 from the 1 in its row. So R2P6 must be one of 2, 3, or 4, which prevents R3P5 from being 3. Hence R3P8 must be 3.
Now look at the 3 in the second row. The 2 in R1P2 blocks R2P1-2. The 4 in R3P4 blocks R2P4-5, the 2 in R3P7 blocks R2P7 and the 3 in R3P8 blocks R2P9. So the only place this 3 can go is in R2P6. The grid thus far:
Continuing in the bottom row:
Let's try to place the 5. As before, R2P2 and R3P3 have to be the same, and neither can be 5 due to the given 5 in R2. In addition, this forces R3P3 to be either 7 or 8. Now suppose R3P5 were 5. Then R1P5 would have to be 7, since the diagonals through R2P5 have the same sum. This then forces R1P3 to be 8, and the same diagonal sums on R2P4 force R1P4 to be 11, a contradiction. Thus R3P2 must be 5.
Hoping this is the breakout:
Let's look at some possible values in the middle of the board. We have the obvious 7 and 8 in R3P3,5. The equal diagonal sums on R2P5 forces R1P5 to be 4 or 5. The equal diagonal sums on R2P4 force R1P4 to be either 7 or 9. Now in the red squares in R2, we have R2P4 is blocked from 1,3,5 in its row, 4 from R3P4, and it also cannot be either 7 or 8, as R3P3 in its diagonal must be one of these values. Hence it is 2, 6 or 9. Similar analysis shows R2P5 must be 2 or 6. The grid thus far:
Picking a winner:
Suppose R1P5 is 5. Then neither R2P4 nor R2P5 can be 6, which forces them to be 9 and 2, respectively. But R2P4 being 9 forces both R1P4 and R3P3 to be 7, which contradicts the equal diagonal sum on R2P4. So R1P5 is 4, which resolves R1P4 = R3P3 = 7, which also gives R2P2 = 7, and R3P5=8. Some progress, but not the breakout I was hoping for...the grid thus far:
The road goes ever onward:
But there is continuing progress! Look at R2P4. From our analysis above, it can only be 2, 6 or 9, and 6 is eliminated by the 7s in its diagonals. Notice that it cannot be 2, due to the constraint that ALL squares with equal diagonal sums are red, and R2P4 being 2 would give equal diagonal sums to R1P4. Thus we must have R2P4 = 9, and the only place the 2 can then go in R2 is R2P5.
Let's finish this off. First note that R1P9 can be only 9 or 10, since 3, 5 and 6 are blocked by its diagonal to the lower left. In the upper left corner, we must have the two diagonals on R3P1 have the same sum, so we must have R1P1 in {3,5,6,9,10} + R2P1 in {4,6,8} add up to 7 + R1P3 in {3,5,9,10} (cannot be 6 due to the 7). Moreover, {R1P1,R1P3} cannot be {9,10}. If R1P3 is 9 or 10, then its diagonal sum is at least 16, which forces R1P1 to be 9 or 10. Hence R1P3 must be 3 or 5. But it cannot be 5, since there is no combination for R1P1,R2P1 that sum to 12. Hence R1P3 must be 3, and (R1P1,R1P2) must be (6,4). The rest of the grid fills in trivially. The final grid:
