Partition a set of numbers into two so that difference between their sum is minimum with equal size

Partition a set of numbers into two such that difference between their sum is minimum, and both sets have equal number of elements. 

For example: {1, 4, 9, 16} is partitioned as {1,16} and {4,9} with diff = 17-13=4. 

Does greedy work here? First sorting, and then picking smallest and largest to fall in set 1, and picking 2nd smallest and 2nd largest to fall in set 2. 

I was asked to prove which I failed :(

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The problem is similar to knapsack problem, however, the number of knapsack is constant in this problem, so :

Observing the insanity of above posts which claim to develop a greedy O(nlogn) solution for a variation of partition problem, I was compelled to code the program. Being NP-hard by nature, the solution falls into pseudo-polynomial time algorithm with complexity O(n^2W) where n = # of elements, W = sum of elements.

//constraints: n is evenvoid fun (int items[], int n){  int total = accumulate (items, items+n, 0) / 2;  int maxSubsetSz = n/2 ;  vector< vector<int> > T (maxSubsetSz+1, vector<int> (total+1, 0) );  //T[i][j] is set if there exists subset of size i that sum up to j  T[0][0] = 1;  for(int i = 0; i < n; i++) //consider taking i-th item     for(int k = maxSubsetSz-1; k >= 0; k--) //each item must be taken once, hence loop backwards      for(int j = 0; j <= total-items[i]; j++)          if ( T[k][j] && items[i]+j <= total )                    T[k+1][j+items[i]] = 1;  for(int j = total; j >= 1; j--)    if ( T [maxSubsetSz][j] ) {      cout << "sum : " << j << endl;       break;    }}