Non-trivial zeros off critical line If non-trivial zeros lay off the critical line (as shown in the picture below),

would they have to come in fours rather than conjugate pairs (as the diagram shows)?
I am presuming they would, since $\sum_{\rho}^{}\text{Li}(x^{\rho})$ is conditionally convergent, and is taken to mean $\sum_{\rho}^{}|\text{Li}(x^{\rho})+\text{Li}(x^{1-\rho})|$, and since $1-(\sigma+bi)$ and its pair only cancel imaginary terms when $\sigma=\frac{1}{2}$, (if $s=\frac{1}{4}+bi$, then $1-s=\frac{3}{4}-bi$), they would presumably have to come in fours, eg:
$s_{1}=\frac{1}{4}+bi$, $1-s_{1}=\frac{3}{4}-bi$, $s_{2}=\frac{3}{4}+bi$, $1-s_{2}=\frac{1}{4}-bi$.
My second question: is it possible for there to be zeros in the critical strip with exactly the same imaginary value?
Update
Just done a bit of playing around & it is easier to visualise how this might happen in a contour plot of $\xi(\sigma+bi)$:

 A: If $s$ is a nontrivial zero of $\zeta$ off the critical line then the four numbers $\{s,\bar{s},1-s,1-\bar{s}\}$ would all be nontrivial zeros off the line. Note $1-\bar{s}$ is the image of $s$ across the critical line, so they are close together, but $\bar{s}$ is the image of $s$ across the real axis, which won't look close to $s$. Your image even depicts one pair of nontrivial zeros at the top with a corresponding pair at the bottom. So yes, the nontrivial zeros off of the critical line come in packages of four.
There are two nontrivial zeros with the same imaginary part if and only if there exists a nontrivial zero off of the line. For if such two zeros existed they can't both have the same real part (else they'd be the same complex number altogether) so one of them must have real part $\ne1/2$, hence it must be nontrivial and off the line. Conversely if there is a nontrivial zero $\rho$ off the line then $1-\bar{\rho}$ will also be a zero and have the same imaginary part.
We don't know if there are nontrivial zeros off of the critical line. This is the subject of the Riemann hypothesis, which is a Millenium problem literally worth a million dollars.
