In an $N$-dimensional real Euclidian space, let an orthant be specified by a vector $\underline{x}_0 = \{x_1, x_2, \dots, x_N\}$ where the components $x_k$ are binary in the sense that $x_k = \pm 1$.
Now let $\underline{x}_j$ be cyclical rotations of that vector $\underline{x}_0$, such that the components are $x_{k+j}$ where the indices are understood modulo $N$. Span a space of vectors $V = {\rm span} \{ \underline{x}_1, \underline{x}_2, \dots, \underline{x}_d \} $. What is the minimum number of $d$ such that $V$ intersects the original orthant indicated by $\underline{x}_0$?
This question can either be understood when notion of the original $\underline{x}_0$ is available, or in an average sense over randomly picked $\underline{x}_0$ where $N$ is large. In the latter case, it is reasonable to make the vector $\underline{x}_0$ almost mean-free, i.e. by imposing $|\sum_{k=1}^N x_k| \le 1$. Also, conditions on $N$ may be useful, i.e. demanding that $N$ is prime.
Background 1: If the vectors $ \{ \underline{x}_1, \underline{x}_2, \dots, \underline{x}_d \} $ were randomly picked and in general position, it is known that, as $N$ turns large, the probability of the span intersecting any particular orthant approaches 1 as $d \ge 0.50 N$ (Thomas Cover, Geometrical and Statistical Properties of Systems of Linear Threshold Devices, 1964). In the present non-random but cyclical case, although the vectors could still be assumed in general position, the probability of the span intersecting the original orthant is, by large scale simulations, known to approach 1 as $d \ge 0.59 N$, i.e. the subspace dimensionality $d$ must be ca. 18 % larger. (M. Schröder, W. Kinzel and I. Kanter: Training a perceptron by a bit sequence: storage capacity, J. Phys. A 29 7965-7972, 1996). To my knowledge, there is no explanation for this published.
Background 2: One could demand a stronger condition, namely that $\underline{x}_0 = \sum_{j=1}^d w_j \underline{x}_j$ holds, with freely chosen $w_j$. However, it can be shown by Discrete Fourier Transformation that this condition can be met only in very special cases, and when $N$ is prime and $\underline{x}_0$ is almost mean-free, it is impossible to achieve. The argument is the following:
Let $\bf{W}$ be a circulant matrix where the generating vector is composed from the components $w_j$, and otherwise zero. Then the condition is equivalent to $\bf{W} \cdot \underline{x}_0 = \underline{x}_0 $. Let $\bf{I}$ be the unit matrix and $\underline{0}$ be a vector of zeros, then equivalently $(\bf{W} - \bf{I})\cdot \underline{x}_0 = \underline{0}$. So $\bf{V} = \bf{W} - \bf{I}$ is also a circulant and we have $\bf{V} \cdot \underline{x}_0 = \underline{0}$. Taking DFTs, $ \underline{\hat x} = {\rm{DFT}}(\underline{x}_0 )$ and $\bf{\Lambda}$ is the diagonal (since $\bf{V} $ is circulant) matrix of eigenvalues of $\bf{V} $. So, after taking DFT, we have $\bf{\Lambda} \cdot \underline{\hat x}= \underline{0}$ . In components, we have that $\lambda_j {\hat x}_j = 0$. Now if $N$ is prime, all DFT components ${\hat x}_j $ are nonzero (for a proof see here), which means that it is required that all $\lambda_j = 0$. This is only possible if all $v_j =0$, in other words, $w_0 = 1$ and all other $w_j = 0$. However, the task has it that $w_0 = 0$, so this condition cannot be met.
If $N$ is not prime, it is still possible that (almost) all DFT components ${\hat x}_j $ are nonzero, so a similar argument holds.
Background 3: Following the discussion above, the span $V$ intersects the original orthant indicated by $\underline{x}_0$ if we define a diagonal matrix $\bf{A}$ with positive elements only, and $(\bf{W} - \bf{A})\cdot \underline{x}_0 = \underline{0}$. So $\bf{V} = \bf{W} - \bf{A}$ is a "circulant plus diagonal", which is disturbing the argument in Background 2. This may be a handle to augment the discussion in Background 2 with a perturbation treatment.