For 8 cards, chosing 3 and leaving 2, there is a translation invariant solution. Arranging the cards based on the differences between them, and the options are:

(1,1,6) eg cards 1,2,3, (2-1 = 1, 3-2 = 1, 1-3 = -2 = 6 (mod 8))

(1,2,5)

(1,3,4)

(1,4,3)

(1,5,2)

(2,2,4)

(2,3,3)

The pair you leave behind is defined by the difference between the second and first card, so it's a number mod 8, but can't be 0, and has to be obtainable as either a number above, or the sum of a number and the number cyclically to its right (eg 1,2,3 I can leave 1,2 for a difference of 1, or 1,3 for a difference of 1+1, or 3,1 for a difference of 6 = -2 mod 8, etc). Many things seem to work, just being greedy requires only 1 correction:

(1,1,6) -> 1

(1,2,5) -> 2

(1,3,4) -> 3

(1,4,3) ->4

(1,5,2) ->7

(2,2,4) ->6

(2,3,3) ->5

And there's lots of room to fiddle, eg 1 and 2 can be switched, as can 3 and 4, 3 and 7, 2 and 6, etc etc. So we should look for a nice way that might generalize to higher n. Will this approach even generalize? We're looking at ordered partitions of n!+k-1 into k positive parts, modulo cyclic translation, and comparing that to ordered k-1 subsets of n!+k-1 modulo translation. Are those even the same size (I'd guess so, but only because I think a translation invariant solution exists)

Not sure if this is going anywhere.