Consider we are given a collection of $n$ vectors in $d$ dimensions, we want to decide if they can be rearranged into $v_1,\ldots,v_n$ such that $\sum_{i=1}^j v_i\geq \textbf{0}$ for all $j\in [n]$. Here $a \geq \textbf{0}$ if and only if all coordinates is at least $0$.
Have this problem been studied before?
I don't know if it has an official name, but it is NP complete; I give a reduction idea from Exact Cover from 3-Sets:
Given $n = 3q$, $X = \{x_1,...,x_{n}\}$ and $C_1,C_2,...,C_m$ a collection of 3 elements subsets.
Use the following set of $3q + 2$ dimensions vectors:
$v_{fill} = [1_1, 1_2, ... ,1_{3q}, \; q_z ,\; 0_b]$ (the subscripts are just labels to show the element "column" position)
For each $C_i = \{ x_{i1}, x_{i2}, x_{i3} \}$
$v_{C_i} = [ 0,...,0, -1_{i1},0,...,0,-1_{i2},0,-1_{i3},0,...,0, \; -1_{z}, \; 1_{b} ]$
$v_{clean} = [ m, m, ..., m, m_{z}, -q_{b} ] $
Starting from $[0,0,...,0,\; 0_{z}, \; 0_{b}]$ the only vector that doesn't contain negative elements is $v_{fill}$ so it must be the first of the sum.
Then in order to be able to add $v_{clean}$ that has a $-q$ in column $b$; we must add exactly $q$ vectors $v_{C_i}$ that contains $[....,1_{b}]$; we cannot add more than $q$ otherwisee the column $z$ becomes negative.
Furthermore we cannot add two $C_i$ that has the same element in the first $3q$ columns, otherwise it would become negative.
At this point $v_{clean}$ can be added; and, finally, the remaining $v_{C_i}$ can be added ("cleaned")
Example:
n = 3q = 6; q=2; m = 3
C_1={1,2,3}, C_2={ 4,5,6}, C_3={1,2,4}
v_fill = [ 1, 1, 1, 1, 1, 1, 2, 0 ]
v_C_1 = [ 1, 1, 1, 0, 0, 0,-1, 1 ]
v_C_2 = [ 0, 0, 0, 1, 1, 1,-1, 1 ]
v_C_3 = [ 1, 1, 0, 1, 0, 0,-1, 1 ]
v_clear= [ 3, 3, 3, 3, 3, 3, 3,-2 ]
Sum: v_fill + v_C_1 + v_C_2 + v_clear + v_C_3
