We can solve this problem using a greedy algorithm. First, sort the presents in non-decreasing order of weight. Then, iterate through the presents while keeping track of the cumulative weight so far. If the current present's weight is not less than the cumulative weight, add it to the stack.

The proof of the greedy algorithm is as follows:

For the sake of contradiction, assume that the greedy solution is not optimal.

The greedy solution selects presents $g_1,g_2,\dots ,g_x$.

The optimal solution selects the lightest presents $o_1,o_2,\dots ,o_y$.

There is a maximal value of $r$ such that $g_1=o_1,g_2=o_2,\dots ,g_r=o_r,g_{r+1}\ne o_{r+1},\dots$

Since $g_{r+1}$ is both a valid choice and has a smaller weight than $o_{r+1}$, $o_{r+1}$ can be replaced by $g_{r+1}$ without worsening our answer. Now, $g_{r+1}=o_{r+1}$ and thus, there is a contradiction in the choice of $o$.

Time complexity: $\mathcal{O}(N\log N)$