This can be solved efficiently via using HashTables.
We can have a hashtable sums sums will store all possible sums of two different elements. For each sum S it returns pair of indexes i and j such that a[i] + a[j] == S and i != j. But initially it's empty, we'll populate it on the way. So, this can be done in O(n^2) time.
Pseudocode
for (int i = 0; i < n; ++i) {
// 'sums' hastable holds all possible sums a[k] + a[l]
// where k and l are both less than i
for (int j = i + 1; j < n; ++j) {
int current = a[i] + a[j];
...
