Solution:

Let the vectors be $\mathbf{x}_i$ and their sum $\mathbf{s}$. Then we have $\mathbf{s} - \mathbf{x}_i = n_i \mathbf{x}_i$ for some scalar $n_i$. Hence $(n_i + 1) \mathbf{x}_i = \mathbf{s}$. If $\mathbf{s}$ is nonzero, then it follows that every vector is a multiple of $\mathbf{s}$ and hence all the vectors are multiples of each other. But we are told that is not true. Hence $\mathbf{s}$ is zero.