This isn't actually Group Theory, but...
(1) It is probably simpler, and not much less accurate (if at all), if you do some rounding. In this case, you could go with 6000 people and 100 groups, say. Or 90 groups. You get the idea.
(2) If I understand the question correctly, you want to have cost levels at leadership levels, and this is independent of groups (I think). You could have the subset sizes as a list, in this case {1000,100,10}, the cost levels a list of the same length (say {3,2,1} in whatever units, hundreds of dollars or whatever). I don't know what specifically you have in mind for officers, but if it is say, chair, cochair, secretary, treasurer, and again there are cost levels of, say, {3,2,1,1}, then we have the needed information at hand.
(3) Now to code it. We'll have three lists as input, corresponding to what is indicated in (2) above (we do not actually need the names of the officers, just how much they cost). We'll do a check that lengths correspond for the community leader sets and their costs, and elements are what we expect (positive integers for sizes of leader sets, nonnegative values for costs). We also want to know how large is the community, and how many groups it has. I'll use "cl" prefixes for "community level" and "g" for "groups" (again, as in subsets of the community, not the groups of group theory).
cost[ccount_Integer, gcount_Integer, clevels_List, clcosts_List,
ocosts_List] /;
ccount > 0 && gcount >= 0 &&
VectorQ[clevels, IntegerQ[#] && # > 0 &] &&
VectorQ[clcosts, Element[#, Reals] && # >= 0 &] &&
VectorQ[ocosts, Element[#, Reals] && # >= 0 &] &&
Length[clevels] == Length[clcosts] :=
gcount*Total[ocosts] + Ceiling[ccount/clevels].clcosts
Pretty simple, albeit not simply pretty...
Anyway, for the example above, I'll show both with and without rounding.
cost[6000, 100, {1000, 100, 10}, {3, 2, 1}, {3, 2, 1, 1}]
(* Out[1241]= 1438 *)
cost[6000, 90, {1000, 100, 10}, {3, 2, 1}, {3, 2, 1, 1}]
(* Out[1242]= 1368 *)
cost[6309, 89, {1000, 100, 10}, {3, 2, 1}, {3, 2, 1, 1}]
(* Out[1243]= 1403 *)
