This problem comes to my thinking inspired from the Cauchy-Schwarz inequality.
Given a vector $u$ in the nonnegative orthant $R^n_+$, what is the largest angle we can get between $u$ and vectors in $R^n_+$? Basically, what is $\min_{v\in \mathbb R^n}{\frac{u^Tv}{\|u\|_2\|v\|_2}}$ given $v\geq 0$?
For example, for $n=2$, it is easy to check that the largest angle is $\arccos{\frac{u_1}{\sqrt{u_1^2+u_2^2}}}$ if $u_1\leq u_2$.
My guess is that the maximum angle is $\arccos{\frac{u_n}{\sqrt{u_1^2+...+u_n^2}}}$ where $u_n:=\min_i {u_i}$ but I cannot show it is true.
I could be wrong! but, inspired by the geometry, I think this problem can be also formulated as follows. Find the vector $v^*\ge 0$ on the surface of the unit $\ell_2$-ball which has the largest distance from $u$. Thus, we need to solve the following: $$ \max_{v\in \mathbb R^n} \ \frac{1}{2}\|u-v\|_2^2\quad \text{ subject to } \quad \|v\|_2^2=1 \ \text{ and } \ v\ge 0. $$ If correct, I think this is a better formulation of the problem and hopefully, one can find its closed-form solution $v^*$. Once that is in hand, the largest angle can be computed. Btw, if it helps, WLOG, we can assume that $\|u\|_2=1.$