My guess would be that Romney will make the 20% threshold as he is at (52 - 16 = 36) 36% now. He may drop some but appears to share the statewide/bonus delegates.
http://www.thegreenpapers.com/P12/KS-R#0310
[the percentage of votes received] = [votes cast for each candidate] ÷ [total votes cast for all candidates meeting the threshold].
Beginning with the candidate who received the most votes, multiply [the percentage of votes received] by [25 National Convention Delegates] and round any remainder up to the next whole number. Repeat for the next highest vote getter until all 25 delegates are allocated. [Section VI. 2. A.]
The 3 party leaders delegates (Todd Tiahrt, the National Committeeman, Helen van Etten, the National Committeewoman, and Amanda Adkins, Chair of the Kansas Republican Party) will attend the convention bound to the candidate who receives the most votes statewide. [Section VI. 3.]
Disregard my 20% threshold post as I misread the 18% Romney number as trailing not as his percentage which it was.
My bad.
This may mean that it would be to Santorum’s advantage to have someone else top 20%, thus cutting the third and fourth place finishers out and giving him the bulk of their delegates. If I am reading the legalese right, If Santorum finishes with 50% and Romney with 20%, the 25 at large would be split 18 to Santorum and 7 to Romney, with everyone else cut out, but if the standings stay as they are, it would be something like 14 to Santorum, 4 to Romney, 4 to Newt and 3 to Paul. By winning by four percentage points less, if the points went to Romney, Santorum would have a net gain of four delegates, and increase his lead on Romney by one.
Of course, I may not be understanding the formula properly.