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Specifically, if $\rho$ is such that $\zeta(\rho)=0$ and $\max_{\rho}\Re(\rho)= \Theta$, can anything interesting be said about the number/distribution of zeros on the vertical line $\sigma=\Theta$?

Clearly this question is almost as hypothetical as they get, so I welcome conditional answers (though not on RH please), consequences of the Bohr-Landau theorem, consequences of the known behavior of $\zeta(s)$ in the critical strip, etc.

Maybe you know something along the lines of ``If there are finitely/ infinitely many, then...''?

I am also interested in why your answer may simply be ``No.''

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Is there any reason (short of RH) to believe that the maximum exists at all? In fact, since there are no zeros with Re=1, this would imply that all zeros have $\Re(\rho)\le1-\epsilon$ for some constant $\epsilon>0$, and this is an open problem. – Emil Jeřábek Jan 10 '13 at 13:52
Good point. I think so: roughly speaking, either there is a maximum (whose value is not known), or there is at least one rogue zero $r$ with $1-\epsilon<\Re(r)<1$, for every $\epsilon > 0$. But then, owing to the known zero-free region, $\Im(r)>T_{\epsilon}\rightarrow\infty$ as $\epsilon\rightarrow 0$. – Kevin Smith Jan 10 '13 at 15:53
That’s not necessarily true either. If there is no maximum, i.e., the supremum is not attained, there is no telling whether the supremum is 1 or smaller. Of course, if the maximum does not exist and $\rho_n$ is any sequence of zeros whose real parts tend to the supremum, then $\lim_n|\Im(\rho_n)|=\infty$, since the roots of any meromorphic function are isolated; this does not need any sophisticated information on zero-free regions. – Emil Jeřábek Jan 10 '13 at 16:08
Obviously - clearly I had implicitly assumed you were arguing about zeros near the line $\sigma=1$, and whether I was discussing an open problem (which is off-topic). Isolation of zeros is not strong enough to prove the statement I made in the above comment. Your original objection is valid - indeed I have not given a good reason to believe a maximum does exist if the supremum is $\leq 1-C$, but I am remaining on topic.. Clearly I need to be more precise here. – Kevin Smith Jan 10 '13 at 17:38
This is getting off topic, but I must say the statement you made above is not very clear to me. Is $T_\epsilon$ some specific function? If not, the following lemma follows easily from the fact that roots are isolated. Let $f$ be a meromorphic function such that $s:=\sup\{\Re(\rho):f(\rho)=0\}$ is finite, and not attained. Then there exists an unbounded nonincreasing function $T\colon(0,+\infty)\to[0,+\infty)$ with the property that $|\Im(\rho)|\ge T(\epsilon)$ whenever $\rho$ is such that $f(\rho)=0$ and $\Re(\rho)\ge s-\epsilon$. – Emil Jeřábek Jan 11 '13 at 11:58

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