A new puzzle type for me! Thanks HATMAN. Haven't looked here in a while so sorry for the delay. Here is my solution and a start to the walkthrough...

791426358
368159742
452378691
527891463
816534927
943267815
285943176
179685234
634712589

1. r6c5 cannot be R - otherwise only 2 R in 27+7 for R27.
=> r8c5 = B. Must be 30/4 extending to r6c8 or r5c7.
Buf if r6c8 = B => Max r6c123 = +12 => Min r7c23+r8c3 = 26.
=> r5c7+r6c567 = B30/4 = {6789}. => r56c8 = R3/2 = {12}

2. r4c5 cannot be R - otherwise only 2 R in 29+5 for R29.
=> r4c5 = B. Must be in 30/4.
Since r5c8 = 1|2 => r45c2 = R3/2 = {12} and r4c345+r5c3 = B30/4 = {6789}

3. In n5 r4c45+r6c56 = {6789}. Also r5c28 = {12} => r5c456 = {345}
=> Max r56c4 = +7 => r5c4 cannot be R => r5c4 = Y. Since r5c4 cannot be 7 Y7 requires at least 2 cells.
R27 requires 4 cells. => r56c4 = Y7/2 = [52].
=> r4c6 = 1, r45c2 = [21], r56c8 = [21].

Max r45c6 = +5 => r5c6 cannot be R => r5c6 = Y.
Since r5c6 cannot be 5 and R29 requires 4 cells => r45c6 = Y5/2 = [14].
=> r5c5 = 3. Also r23c56 = R29/4 = {5789}.

4. Since r5c7 is B => B in 22+13 must be in r2c7 or r4c9 or r2c7+r3c78.
Must be latter since both 22+13 require at least 2 cells. => r4c789 = G.
Since r4c345 = 3 of {6789} => Max r4c789 = 9+5+4 = +18. => r4c789 = G13/3. Must be {346}.
=> r5c3 = 6, r4c345 = {789}, r4c1 = 5, r6c9 = 5.
Also => r2c7+r3c78 = B22/3 => Includes a 9.

Since r5c3 is B => B in 22+16 must be in r6c1 or r8c3 or r7c23+r8c3.
Must be latter since both 22+16 require at least 2 cells. => r6c123 = G.
Since r6c567 = 3 of {6789} and 5&6 already in n4 => Max r6c123=9+3+4 = +16.
=> r6c123 = G16/3 = {349} => r5c6=9 and r6c567={678}

5. 11+10 each require 2 cells. => r89c2 = R10/2 and r9c34 = Y11/2.
=> Hidden single r9c5=1.
=> In n8 2 must be in r789c6. (9 in n7 must be in G22/3)

6. In B22/3 @r2c6 r3c8=9. => 9 cannot be anywhere in 9+9 => r1c67 = 9/2 and r12c8 = 9/2.
r1c6 can only be 3 or 6. => r1c67 = {36} => r12c8 = {45}
=> r23c7={67} => r1c67=[63] => r123c9 = {128}

7. Since {124} are already in c8 the 7+7 cannot be 7/1 + 7/3 => Must be 7/2@r8c6 + 7/2@r8c8.
Only possibility for r8c89 = [34].
3 cannot go in r9c5 or r9c8 => 13+3 = 3/2@r9c5 + 13/2@r9c7. => r9c6=2 and r9c78=[58]

8. In R29/4 9 must go in r2. => 9+9 must be 9/2@r2c1 + 9/2@r2c3.

Just mopping up from here.