Are Li's numbers strictly increasing? Li's numbers $\lambda_{n}$ are defined by:
$\lambda_{n} = \sum_{\rho} \Big(1-\Big(1-1/\rho\Big)^n + 1-\Big(1-1/{\overline{\rho}}\Big)^{n}\Big)$
Where $n$ is real and $\rho$ runs over all the nontrivial zeros of the Riemann zeta function and  is always paired together with $1-\rho$.
We claim that the derivative of $\lambda$ with respect to $n$ exists everywhere in $\mathbb{R}$.
Indeed, begin with
$\frac{d}{dn} \lambda_{n} = -\sum_{\rho} \Big(\Big(1-1/\rho\Big)^n\log\Big(1-1/\rho\Big) + \Big(1-1/{\overline{\rho}}\Big)^n\log\Big(1-1/\overline{\rho}\Big)\Big) $
Then note that for any fixed real $n$, as Im$\rho \to \infty$, the summand tends to $-\Big(1/\rho + 1/\overline{\rho}\Big)$, and convergence occurs due to the pairing of the zeros.
We therefore conclude that $\lambda$ is differentiable (hence continuous) everywhere in $\mathbb{R}$.
Now, let $h(n) = \lambda_{n} + \lambda_{-n}=2\lambda_{n}$.
$h(n)=:$
$\sum_{\rho}  \Big(1-\Big(1-1/\rho\Big)^{n} +  1-\Big(1-1/\overline{\rho}\Big)^{n} +  1-\Big(1-1/\rho\Big)^{-n} +  1-\Big(1-1/\overline{\rho}\Big)^{-n}\Big) + \sum_{1-\rho}  \Bigg(1-\Bigg(1-\frac{1}{1-\rho}\Bigg)^{n} + 1-\Bigg(1-\frac{1}{1-\overline{\rho}}\Bigg)^{n}  +  1-\Bigg(1-\frac{1}{1-{\rho}}\Bigg)^{-n} +  1-\Bigg(1-\frac{1}{1-\overline{\rho}}\Bigg)^{-n}\Bigg) $
Observe that in the above, the sum over $\rho$ is exactly equal to the sum over $(1-\rho)$, so that for the purposes of this argument, it suffices to consider only the sum over $\rho$ forthwith.
Suppose that $h$ has a stationary point at some positive real $m$, so that $h'(m)=0$: 
$h^{'}(m)=:$
                                                                                                         $\sum_{\rho}\Bigg(-\Bigg( \log\Big(1-1/\rho\Big)^{\Big(1-1/\rho\Big)^{m}}  +  \log\Big(1-1/\overline{\rho}\Big)^{\Big(1-1/\overline{\rho}\Big)^{m}} \Bigg) +                                                                                                   \Bigg( \log\Big(1-1/\rho\Big)^{\Big(1-1/\rho\Big)^{-m}}  +  \log\Big(1-1/\overline{\rho}\Big)^{\Big(1-1/\overline{\rho}\Big)^{-m}} \Bigg)\Bigg)$
Which simplifies to
$h^{'}(m)=\log\prod_{\rho} \beta_{\rho}$
Where
                                                                           $\beta_{\rho} = \Big(1-1/\rho \Big)^{\Big(1-1/\rho \Big)^{-m} -\Big(1-1/\rho \Big)^{m}}\Big(1-1/\overline{\rho} \Big)^{\Big(1-1/\overline{\rho} \Big)^{-m} -\Big(1-1/\overline{\rho }\Big)^{m}}$
Therefore, if $h^{'}(m)=0$, we should have
                                                                                                       $\prod_{\rho} \beta_{\rho}=1$
But one can quickly verify that this is impossible for any nonzero real $m$, and a contradiction is reached.
Hence it follows that $h$ has no stationary points for $n>0$, and by its continuity, this implies that $\lambda_{n}$ is either a strictly increasing or strictly decreasing function of $n$ for $n> 0$. But we already know from computational evidence  that $0<\lambda_{1} < \lambda_{2} < \lambda_{3}<\cdots<\lambda_{3 000} $, therefore the only remaining possibility is that    $\lambda_{n}$ is strictly increasing with positive $n$ ?
 A: I think positivity of $\lambda_n$ is Li criterion, which is equivalent 
to Riemann hypothesis.
See A Li-type criterion for zero-free half-planes of Riemann's zeta function
A: Li's criterion is very well explained in joro's linked article.
It applies to any entire function $\prod_\rho (1-\frac{z}{\rho})(1-\frac{z}{\overline{\rho}}), \rho \ne 1$ as a criterion that $Re(\rho) \le 1/2$ (*),
and so does your "non vanishing of $\frac{d}{dn}\lambda_n$" idea.
Hence, for proving the RH you have to replace "one can quickly verify that this is impossible for any nonzero real $m$" by a deep argument making $\xi(s)$ special, and that doesn't apply to, say, $\xi(s)(s-a)(s-\bar{a}), Re(a) > 1/2$.
Take a look at the Selberg class, you'll see all the properties of $\zeta(s)$  you need to use in order to hope proving the RH (once you remove any of those properties, you get many functions very close to $\zeta(s)$ but for which the RH fails).

(*) Li's criterion is indeed very simple : $|1-\frac{1}{\rho}| < 1 \Leftrightarrow |\rho-1| < |\rho| \Leftrightarrow Re(\rho) > 1/2$.
Then, starting with an entire function $f(s) = \prod_\rho (1-\frac{s}{\rho})$, let $\phi(z) = f(\frac{1}{1-z})$.
Its zeros are at $\frac{1}{1-z} = \rho \Leftrightarrow z = 1-\frac{1}{\rho}$. Hence, $\frac{\phi'(z)}{\phi(z)}$ is holomorphic on the unit disk $|z| < 1$ if and only if $|1-\frac{1}{\rho}| \ge 1$ i.e. iff $Re(\rho) \le 1/2$.
Write the Taylor series $\frac{\phi'(z)}{\phi(z)} = \sum_{n=0}^\infty c_n z^n $ and since $\log \phi(z) = \sum_\rho \log (1-\frac{1}{\rho (1-z)}) = \sum_\rho \log(1-\frac{1}{\rho}-z)-\log(1-z)$ we have $$\frac{\phi'}{\phi}(z) = \sum_\rho \frac{1}{1-z}-\frac{1}{1-\frac{1}{\rho}-z}$$ so that $c_n = \sum_\rho 1-(1-\frac{1}{\rho})^{-n-1} = \sum_\rho 1-(1-\frac{1}{1-\rho})^{n+1}$. 
The condition $Re(\rho) \le 1/2$ then becomes equivalent to $\lim \sup_{n \to \infty} |c_n|^{1/n} \ge 1$.
Finally, if $\rho$ comes with its complex conjugate $\bar{\rho}$, then $c_n = \sum_{\rho} Re(1-\Big(1-\frac{1}{1-\rho}\Big)^{n+1})$ that is obviously $\ge 0$ for any $n \ge 0$ whenever $|1-\frac{1}{1-\rho}| \le 1 \Leftrightarrow|1-\frac{1}{\rho}| \ge 1 \Leftrightarrow Re(\rho) \le 1/2$ . Conversely, ordering the $\rho$ by $|1-\frac{1}{1-\rho}|$, you see that the largest $|1-\frac{1}{1-\rho}|> 1$ term dominates the sum  as $n \to \infty$, and hence $c_n$ becomes negative sometimes .
