Clarification of Answer... find the max possible two equal sum in a SET

I think thinking how this solution relates to the single subset problem might be misleading for you. Here we are concerned with a maximum achievable sum, and what's more, we need to distinguish between two disjoint sets of numbers as we traverse. Clearly tracking specific combinations would be too expensive.

Looking at the difference between sets A and B, we can say:

A - B = d
A = d + B

Clearly, we want the highest sum when d = 0. How do we know that sum? It's (A + B) / 2!

For the transition in the dynamic program, we'd like to know if it's better to place the current element in A, B or neither. This is achieved like this:

e <- current element
d <- difference between A and B

(1) add e to A -> d + e

why? 
A = d + B
(A + e) = d + e + B

(2) add e to B -> d - e

why? 
A = d + B
A = d - e + (B + e)

(3) don't use e -> that's simply
  what we already have stored for d

Let's look at Peter de Rivas' code for the transition:

# update a copy of our map, so
# we can reference previous values,
# while assigning new values
D2=D.copy()

# d is A - B
# s is A + B
for d,s in D.items():

  # a new sum that includes element a
  # we haven't decided if a 
  # will be in A or B
  s2 = s + a

  # d2 will take on each value here
  # in turn, once d - a (adding a to B),
  # and once d + a (adding a to A)
  for d2 in [d-a, d+a]:

    # The main transition:
    # the two new differences,
    # (d-a) and (d+a) as keys in
    # our map get the highest sum
    # seen so far, either (1) the
    # new sum, s2, or (2) what we
    # already stored (meaning `a`
    # will be excluded here)
    # so all three possibilities
    # are covered.
    D2[abs(d2)] = max(D2[abs(d2)], s2)

In the end we have stored the highest A + B seen for d = 0, where the elements in A and B form disjoint sets. Return (A + B) / 2.