Let n >= 3 be an integer. Prove that there exists a set S of 2n positive
integers satisfying the following property: For every m = 2,3,...,n
the set S can be partitioned into two subsets with equal sums of elements,
with one of subsets of cardinality m.

Let's try n = 3.
So it can be partitioned into 2 + 4, or 3 + 3,
the 2 are probably the biggest two elements.

I don't know, 9+10+11+12 = 19 + 23.
Can I partition it now into 3 + 3?
  19 + (11+12) = 23 + (9+10)
  not surprising as the elements are chosen so that
  a = b+c, e = d+f
  so there are partitions
  a+e = b+c+d+f
  and
  a+d+f = b+c+e

Maybe, the n=3 was too easy.

n=4.
Again, let's have
a = a1+a2+a3
b = b1+b2+b3
So
a+b = a1+a2+a3 + b1+b2+b3
and
a + b1+b2+b3 = b + a1+a2+a3.

But I also need a 3+5 partition.
a + a1+a2 = a3+b + b1+b2+b3 = a3 + 2b
2a - a3 = a3 + 2b
2a = 2a3 + 2b
a = a3 + b
a1+a2 = b
seems doable.

So b1,b2,b3 = 1,2,3
b = 6,
a1,a2, ... oops, I cannot make it so a1+a2 = 6 and they both differ
from b1,b2,b3.
But it is not so essential that they are integers,
if they were fractions, I can just multiply them all by the common denominator.
So it is possible to choose a1,a2 so that a1+a2 = b1+b2+b2.

But in general, it looks messy.
Maybe it is really a good start to choose the partitions.

There is the partition 2+6, so there are x1+x2 = x3+x4+...+x8
The 4+4 partition is symmetrical, and none of the parts can contain both
x1+x2, so it is really WLOG:
x1+x3+x4+x5 = x2 + x6+x7+x8.
Or maybe, it would be better to start with the 3+5:
again, none of the parts can contain both x1,x2, so
it is WLOG (ignoring the 4+4 above)
x1+x3+x4 = x2+x5+x6+x7+x8

So, basically, there must be two partitions on
x3, ..., x8, one of the form 2+4, the other of the form 3+3
such that in one partition, the groups are not equal
but the difference is the same as in the other partition.
If it was supposed to be equal, I would get back the original problem and could use induction.
But it cannot be equal since it would mean x1=x2,
contradiction with them being different members of a set.

But if I find the partition where they are equal, I can raise
a single element by something to make the partitions equally unequal.
And use them in the next partition.

The problem seems to be solved, now to write it down in detail.
The problem is solved by induction on n, for n=3, we have e.g.
  9+10+11+12 = 19 + 23.
  23 + 9+10 = 19 + 11+12.
Now, for a given set satisfying the property, we construct another set
for n one bigger.
Let X be big enough (something like bigger than the sum of all the elements of the previous set)
Then we add X to any of the numbers.
And add two new numbers: 2X, 3X.
By the choice of X, all the numbers are still different.
(now I see that probably just adding 1 to the biggest number in the original
set, and adding two new numbers X, X+1 would probably also have done the job,
but let's keep the original version).

Wait, 2X, 3X was not the right choice,
I need two numbers a,b so that:
a-b = X       -- for converting older solutions
a+b = X + sum -- for the 2+(n-2) decomposition
2b = sum
2a = 2X+sum
The sum must be even, but it actually is as the original decomposition were possible.
b = sum/2
a = X+sum/2
What if sum/2 already is in the set?
Can it even be?
  If there was some x in the set of the size sum/2, then the only balanced decomposition
  is x = the_rest_of_the_set.
  Contradiction.

So we know that b was not in the set.
Then we set the X to let's say the sum of the original solution.
So b = X/2.
a = 3X/2
Let's call the raised element from the original set c (its original value).
So now, instead of c, we have c+X.
Then a=3X/2 > c+X > b=X/2, so we still have a set.

Any original partition will now differ by X,
so we appropriatelly add 'a' and 'b' to it to nulify the problem.
And for the 2+(n-2) decomposition, we use that a+b = 2X = the_rest.

Problem solved!