It is well known that whenever $\rho$ is a nontrivial zero of the Riemann zeta function $\zeta(s)$, then $1-\rho$ is also a zero.
But does the equality $\Re \sum_{\rho} \dfrac{1}{\rho} = \Re \sum_{\rho} \dfrac{1}{1-\rho}$ hold, where each zero is paired with its conjugate and $\rho$ runs over the entire set of nontrivial zeros of $\zeta(s)$ ?
If the conjugate of $\rho$ is $1-\rho$, and if by $$\sum_\rho \frac{1}{\rho}\tag{1}$$ you mean to pair each zero with its conjugate, then by $(1)$ you really mean $$ \sum_{\rho} \frac{1}{\rho} + \frac{1}{1 - \rho}.$$ This is precisely the same as your second sum, which is precisely $$ \sum_{\rho} \frac{1}{1 - \rho} + \frac{1}{\rho}.$$ So your two sums are exactly the same sum.