1 is not the loneliest number:
Start off by shading all places where a 1 cannot go, either by Sudoku, anti-knight, anti-king, or non-consecutive with 2. There's actually quite a bit:
We see that box 6 in each grid only has one possible location for the 1, and we're off. Note the clever exclusion of 1 from the upper grid of R9C4 using the anti-knight rule with the 1 in the lower grid (R4C8). We're able to place all ones in the upper grid, but the lower grid requires a bit more work:
PATCH
As pointed out by @fljx (thanks!), in my original version of the solve, I mistakenly excluded 1 from R9C5 of the lower grid by faulty application of the non-consecutive rule. Got lucky (and unlucky) that it lead to the answer. I normally wouldn't grind this way, but there's a lot of text after here that depends on R9C5 not being 1, so I'm just going to case this away. Sorry.
Assume R9C5 is 1. Then the 1 in box 2 must be in R2C4 or R3C4, which forces the 1 in box 1 to be in column 1. This looks down to box 7, and we readily find the other 1s in R2C4. R3C1, R5C3 and R8C2.
Working in the lower grid, we can now try to place 2s. Box 3 is the key, since the 2 in the upper grid, and the non-consecutive rule forces 2 to be in R3C7. Box 9 is an easy win too, since the 2 in R8C6 forces R9C9 to be 2. We can then place the 2 via singles in box 6, box 2, then box 5, and the rest fall. Our hypothetical grid, with non-candidate locations for 3 shaded:
R4C6 cannot be 3, as it would block box 2. Now R7C5 cannot be 3, as it would block box 5. This forces the 3 in box 8 to be in column 4, which removes R6C4 from consideration. Now R3C5 cannot be 3, as it would block box 5, forcing R2C6 to be 3. This quickly forces R3C9 to be 3, and then R5C7, but this prevents any cell in box 5 from being a 3, a contradiction. Hence R9C5 cannot be 1. Ugh. Really, really sorry about that.
Back to the original solve:
At this point the 1 in box 8 of the lower grid must be in column 4, forcing the 1 in box 2 to be in R3C5, which prevents the 1 in box 1 from being in column 3. Now a little more clever, we know one of R8C4 or R9C4 is a 1, so neither of R8C3 nor R9C3 is a 1 by anti-king, and neigh R8C2 nor R9C2 is a 1 by sudoku/anti-knight. This forces the 1s in boxes 1 and 4. We are left with two possible placements of 1s in boxes 7 and 8.
Now do 2s:
Proceed the same way, shading cells that cannot be 2 by the ruleset:
The 2 in the shared box must be in column 1 of the upper grid, forcing R3C2 of the upper grid to be 2, which forces R4C5 to be 2. One of R7C6 and R8C6 must be 2, so none of R7C&, R7C8, nor R8C8 can be 2, meaning the 2 in box 9 is in column 9. This forces the 2 in box 6 to be in R5C8. Now one of R1C7 or R2C7 is 2, meaning R1C6 cannot be, so all of R1C4, R2C7 and R7C6 are 2, leaving two possible positions for the 2s in boxes 7 and 9.
In the lower grid, the only place for 2 in box 9 is R9C9. This forces the 2 in box 7 to be in row 7, which means the 2 in box 4 cannot be in row 6. The 2 in box 3 must be in row 7, forcing the 2 in box 6 to be in R6C8. Let's work in box 1 a little bit. The 2 in box 2 is in one of R1C4, R1C5 or R2C4, all of which see R1C3 by sudoku or anti-king rules. Similar logic blocks 2 from R3C1.
But now I see where I forgot to mark a blocked 2 in box 3 by non-consecutive rules, so R3C7 is 2. This lets us place the rest of the 2s with straightforward sudoku logic. The grid thus far:
For 3s, non-consecutive is going to have to get us started:
Locations where 3 cannot be:
Notice that in the upper grid, R9C3 cannot be 3, since a 3 there would block all candidates for a 3 in box 8. This forces the 3 in box 1 of the lower grid to be in row 3. Now a 3 in R2C6 of the lower grid would block a 3 from going in box 3 of the lower grid, forcing the 3 in box 3 of the upper grid to be in row 2. But this is the same as box 7 of the upper grid, and this 3 looks right forcing the 3 in box 9 of the upper grid to be in column 8, and we get some candidate eliminations looking up. But notice that R8C4 of the upper grid cannot be 3, since it would block both candidates in box 7. This gets us our first 3 placement in R7C8 of the upper grid, which immediately forces R4C7 to be 3. R3C4 is now the only location in row 3 that can be a 3, and this placement forces all the 3s in the upper grid, except for two candidates in boxes 1 and 3. In particular, we now have a 3 in box 7, which we can use to attack the lower grid:
In the lower grid, the 3 in box 3 looks down and forces the 3s in boxes 6 and 9. The 3 in box 4 must now be in row 6, which looks right and forces the 3s in boxes 5 and 1; the latter looks back down to box 4, and all 3s in the lower grid are easily placed:
On to 4s:
Grid with non-candidates shaded:
We get two immediate placements in box 5 of the upper grid and box 4 of the lower grid, which both look down forcing the 4s in the box below. Now in box 9 of the upper grid, the 4 must be in row 9, looking left to force R7C2 to be 4, which forces R4C1 to be 4, and thus R6C7. In the lower grid, box 8's 4 must be in row 9, forcing R7C7 to be 4, which with anti-knight looks back to forces R9C8 to be 4. There is now only one place for 4 in box 2, and its forces all 4s in the lower grid. Back in the upper grid, the situation is a bit messy, but we can remove R2C8 as a candidate, since it would block a 4 from going in box 2. Moreover, neither R1C9 nor R2C9 can be a 4, because then the 3 in that box would be consecutive with it. Current state:
5s:
Our candidates...surprisingly limited!
We get some immediate placements: lower grid box 8 (which also finishes our 1s...huzzah!) and box 4. The former looks right to force the 5 in box 9 to be in row 9, which forces box 7. This 5 looks up to box 1, and with anti-king on the 5 in box 4 we get R1C2 is 5, and then easy sudoku places the rest of the 5s in the lower grid. In the upper grid, we have singular candidates in boxes 4 and 5, and combined with the shared box 7, the 5 in box 1 is forced to R3C3, which resolves the 4s in box 1 by non-consecutive. We can then chase this 4 around to finish their placement. Using non-consecutive with 4 in box 9 forces R8C7 to be 5, and the rest of the 5s fall easily. The grid thus far:
6s:
Candidates:
We have some singular placements: upper grid box 7 and row 7, lower grid box 8 and row 8. In the lower grid, we force R5C8 to be 6, which looks left, and with anti-king from R7C6 forces the 6 in box 5 to be in R4C5. This looks left to resolve box 4, with anti-king resolves box 2, and with anti-knight resolves box 1. Sudoku finishes the lower grid.
In the upper grid, the 6 in R8C1 looks up, forcing R2C2 to be 6, and thus R5C3. With anti-king and sudoku, this looks right to force R6C6 to be 6, and resolves box 6 as well. The remaining 6s are placed by straightforward sudoku.
Finishing up:
Candidates for 7s:
The noose is closing in, and it is basically fill-in:
To finish up, we note that 8s cannot be placed next to 7s, which we really only need once, and sudoku finishes the puzzle:
