I have the following data sheet

1 2 3 4 5
6 7 8 9 0
0.1 0.2 0.3 0.4 0.5
0.6 0.7 0.9 0.8 0.1

I would like to implement a conditional formatting formula that would:

  • bold the maximum number of each row (i.e. I1, H2, I3, G4 would be bold);
  • bold+red the maximum number of each block, where a block consists of multiple rows - E1:I2 is one block, E3:I4 is another, etc. (i.e. H2 and G4 would be red + bold).

The first rule is easy to implement. I simply apply the condition =E1=MAX($E1:$I1) on the entire range E:I.

For the second rule, what I currently have is the condition =E1=MAX($E$1:$I$2) on the range E:I, however this only works for the first block. One way to solve it is to apply it to the range E1:I2 instead, then copy the same formula to the range E3:I4, manually fix the row numbers in the condition, and repeat for every block below. However, there are quite a few blocks below, so this solution isn't ideal. Is there a better way of implementing this blockwise formatting?


A note on @doubleunary's accepted answer below - the answer only works for blocks of size 2 (i.e. 2 rows). For larger blocks (in my case of size 6) I had to use the following, more general formula (fortunately none of my cells are empty so I didn't have to worry about that part):

=E1=MAX(OFFSET($E1:$I6, -MOD(ROW(E1) - 1, 6), 0))

1 Answer 1


Use offset(), row() and iseven(), like this:

=isnumber(E1) * ( E1 = max(offset($E1:$I2, -iseven(row(E1)), 0)) )

The isnumber() test prevents the conditional format from highlighting blank rows.

  • excellent :) . However, you must add a filter for the cells of the range that are empty, like: =and (E1=max(offset ($E1:$I2, -iseven (row (E1))), 0)), E1<>"")
    – Daniele
    Oct 13, 2022 at 15:16
  • Thanks. Edited the answer. Oct 13, 2022 at 15:59
  • Thanks for this. Offset was indeed the key idea here, though I had to adapt this answer a little bit for my use-case (since in my actual, much larger datasheet, the blocks were actually bigger than 2 rows). I'm accepting this answer but editing my final solution into the question for posterity. Oct 13, 2022 at 22:05

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