Reduce set partition search to decision?
The standard approach works fine to reduce the problem size.  Given a "true" instance, $n$ elements which can be partitioned, you want to be able to reduce it to a smaller sized problem.  Grab any 2 elements, and combine them into one (having their sum as its size), and now you have an instance with $n-1$ elements.  If the answer to that instance is yes, great.  If it is no?  They must go into different partitions, which (effectively) adds their difference to whichever partition got the larger one.  Get rid of both, but add the difference as a new element.  In either case, you have $n-1$ elements in your new instance.