Implications from the puzzle rules:

(I1) A 1 cannot be on any cell with a Red Mark (RM), or a 9 in any cell with a Green Mark (GM). A 2 can have at most one RM (and that with a 9), etc.

(I2) Since the puzzle is NC - each GM must have at least one of (123) in the cells it straddles.

Similarly a RM must have at least one of (789) in the cells it straddles.

(I3) A Blue Mark (BM) must have one number smaller than 5 and one number larger than 5 in the cells it straddles.

1. n9

of the numbers (789)...

(I2) -> One in r89c7, one in r8c89

(I2) -> One in r7c8

(I3) -> One in r8c9

-> r8c8 and r9c9 both from (123)

-> r9c8 not from (123)

(I1) -> HS 1 in n9 -> r9c9 = 1

-> r8c9 = 9

-> HS 2 in n9 -> r8c8 = 2

-> r7c8 = 8

-> 7 in r89c7

-> 3 in r7c79

2. n8

(I1) -> HS 1 in n8 -> r7c5 = 1

(I2) One of (789) in r78c4, one in r78c6

(I2) -> one of (78) in r9c5. (Cannot be 9 since r9c56 cannot be [91])

-> 9 in r7c46

-> HS 2 in n8 -> 9c6 = 2

-> r9c5 = 8

3. n6

(I1) -> 1 in r4c78

8 in r7c8 -> r6c8 not from (789)

(I2) -> r6c9 from (78)

-> r5c9 not from (789)

(I2) -> r5c8 from (789). Only possibility is 7.

-> r56c8 = [73]

-> r6c9 = 8

(I1, I2) -> r5c7 = 9

r5c9 at least 4 -> (I2,NC) r4c79 = [12]

(NC) -> r6c7 = 6

-> r4c8,r5c9 = {45}

4. n3

(28) both in r123c7 -> r12c7 = {28}

(I1) -> 1 in r12c8 (adjacent to the 8)

(I1) -> r3c8 = 9

Since r3c9 is max 7 -> r2c9 is min 3

-> r1c9 is max 6

Also r1c9 is min 3 -> r2c9 is max 6

-> HS 7 in n3 -> r23c9 = [37]

5. c123

(I1) -> r8c3 = 8

(I1) -> r9c2 = 9

Also (I1, NC) -> r4c13 = [89]

-> r5c3 = 1

(I2) One of (789) in each of r12c1, r12c3, and r3c2

They can only be in that order 9,8,7

(I1) -> r1c2 = 1 -> r1c1 = 9

Also HS r6c2 = 7

6. Some more...

HS r2c8 = 1

-> r12c7 = [28]

-> (NC) -> r1c89 = [46], r3c7 = 5

-> r1c45 = {37}

-> r2c3 = 7

-> r1c36 = [58]

Also r9c8 = 6

-> r8c7 = 7, r7c9 = 5, r9c7 = 4

-> r4c8 = 5, r5c9 = 4

etc.