How to efficiently find positions of duplicates?

You can use GatherBy for this. You can map List onto Range[...] first if you wish to have exactly the same output you showed.

positionDuplicates[list_] := GatherBy[Range@Length[list], list[[#]] &]

list = {3, 3, 6, 11, 13, 13, 11, 1, 2, 3, 12, 8, 9, 9, 4, 15, 5, 6, 9, 12}

positionDuplicates[list]

(* ==> {{1, 2, 10}, {3, 18}, {4, 7}, {5, 6}, {8}, {9}, 
        {11, 20}, {12}, {13, 14, 19}, {15}, {16}, {17}} *)

If you prefer a Sow/Reap solution, I think this is simpler than your version (but slower than GatherBy):

positionDuplicates[list_] := Last@Reap[MapIndexed[Sow[#2, #1] &, list]]

If you need to remove the positions of non-duplicates, I'd suggest doing that as a post processing step, e.g. Select[result, Length[#] > 1&]

Here is a version based on sorting, and using Mr. Wizard's dynP function:

dynP[l_, p_] := 
   MapThread[l[[# ;; #2]] &, {{0}~Join~Most@# + 1, #} &@Accumulate@p]

positionOfDuplicates[list_] :=
   With[{ord = Ordering[list]},
      SortBy[dynP[ord, Length /@ Split[list[[ord]]]], First]
   ]

so that

positionOfDuplicates[list]

(* {{1,2,10},{3,18},{4,7},{5,6},{8},{9},{11,20},{12},{13,14,19},{15},{16},{17}} *)

It is also fast enough, although not as fast as the one based on GatherBy.

If you wanted to retain each value as well as its positions, this works.

Sort[
 Map[{#[[1, 1]], Flatten[#[[All, 2]]]} &, 
  Reap[MapIndexed[Sow[{#1, #2}, #1] &, list]][[2, All, All]]]]

(* Out[178]= {{0, {14}}, {1, {17, 19}}, {4, {4, 
   20}}, {5, {12}}, {7, {10}}, {9, {13}}, {10, {2, 
   6}}, {11, {3}}, {12, {7, 15}}, {13, {8, 9, 11}}, {14, {1, 16, 
   18}}, {15, {5}}} *)

It's maybe 20x slower than the GatherBy though.