Given rank-$1$ square matrices $A_1, A_2, \dots, A_n$, determine if there exists $x \in \mathbb R_{>0}^n$ such that
$$ \sum_{i=1}^n x_i A_i = x_1 A_1 + \cdots + x_n A_n $$
is singular, or decide whether none exists (i.e., the positive linear combination is nonsingular for all positive $x$). Is there a well-known method or algorithm to do this?
Here is a partial answer, and perhaps someone will be able to complete it.
Suppose that $A\in\mathbb{R}^{m\times m}$. Because each $A_i$ is rank 1, it can be written as a dyad $A_i=f_i g_i^T$, where $f_i$ and $g_i$ are vectors. Then $$\sum_i x_i A_i = \sum_i x_i f_i g_i^T = F X G^T$$ where $X\triangleq\mathop{\textrm{diag}}(x)\in\mathbb{R}^{n\times n}$, and $F,G\in\mathbb{R}^{m\times n}$ collect the vectors $f_i$, $g_i$, respectively, as columns. Let $$r_F\triangleq\mathop{\textrm{rank}}(F)\leq\min\{m,n\}, \quad r_G\triangleq\mathop{\textrm{rank}}(G)\leq\min\{m,n\}.$$ Since $X$ is full rank, then we have $$\mathop{\textrm{rank}}(FXG^T)\leq\min\{r_F,r_G\}\leq m$$ So if either $F$ or $G$ has a rank of less than $m$, you're done. In particular, this implies that $n\geq m$ if this is going to have a non-trivial answer.
If not, you have more work to do. And that's where I am personally stuck.
Let $(U_F,\Sigma_F,V_F)$ and $(U_G,\Sigma_G,V_G)$ be economy-sized SVDs of $F$ and $G$, respectively. This means that $$U_F\in\mathbb{R}^{m\times r_F} \quad \Sigma_F\in\mathbb{R}^{r_F\times r_F} \quad V_F\in\mathbb{R}^{r_F\times n}$$ $$U_G\in\mathbb{R}^{m\times r_G} \quad \Sigma_G\in\mathbb{R}^{r_G\times r_G} \quad V_G\in\mathbb{R}^{r_G\times n}$$ Then $$FXG^T=U_F\Sigma_FV_F^TXV_G\Sigma_GU_G^T$$ It's not difficult to see that the unknown here is the rank of the $r_F\times r_G$ matrix $V_F^TXV_G$. Note that this is not an SVD, though it looks almost like one. Indeed, it is just a (possibly) reduced version of the very problem you began with!
I know this isn't a complete answer but I figured I shouldn't let the effort go to waste. If someone else can be inspired on this to finish the task, even the OP, then by all means, I look forward to voting them up.