My question is simple: can we show that the sum
$$ \sum_{k=1}^{p-1} \frac{\left( \frac{k}{p} \right)}{k} $$
is positive for all primes $ p $ where $ (k/p) $ denotes the Legendre symbol modulo $ p $, i.e. equal to $ 1 $ if $ k $ is a quadratic residue modulo $ p $ and equal to $ -1 $ otherwise? After dividing this by $ H_p $ we can see this as a kind of "logarithmic average" of the Legendre symbol over $ \mathbf Z/p \mathbf Z $.
If we ask that the sum be infinite, then we can show this by apealing to the standard result about the nonvanishing of $ L(1, \chi) $ for $ \chi $ a nontrivial Dirichlet character. In this case we could possibly even exert greater control over the $ L $-function by linking it to the Dedekind zeta function of a number field, for example.
However, passing from what we might know about $ L(s, \chi) $ to a bound on this truncated sum seems nontrivial. It's possible that there's a purely elementary argument that would do the trick here, but if there is I haven't been able to find it in a short time, so I've decided to ask the question here instead.
Note: I've confirmed by computer search that the sum in question is indeed positive for all primes less than $ 2000 $. However, this doesn't necessarily mean much, since the average in question is connected to the Liouville function whose logarithmic mean has a strong bias to be positive due to $ \zeta(1/2) $ being negative.
In particular, any attempt to construct a possible completely multiplicative sign sequence to give a counterexample runs afoul of this problem, so even if the statement doesn't hold, the first counterexample prime $ p $ may be quite large.
Best Answer
Consider the infinite sum $$ \sum_{k=p}^{\infty} \frac1k \left( \frac kp \right). $$ Polya Vinogradov inequality implies that for $$ |A(t)|=\left| \sum_{k\leq t} \left( \frac kp \right)\right|< p^{1/2} \log p. $$ By partial summations, we have $$ \sum_{k=p}^{\infty} \frac1k \left( \frac kp \right)=\int_{p-}^{\infty}\frac{dA(t)}t = \frac{A(t)}t \bigg\vert_{p-}^{\infty} +\int_{p-}^{\infty} \frac{A(t)}{t^2} dt. $$ Thus, we have $$ \left| \sum_{k=p}^{\infty} \frac1k \left( \frac kp \right)\right|\leq \frac{2\log p}{p^{1/2}}. $$ The sum of our interest is $$ L(1,\chi)-\sum_{k=p}^{\infty} \frac1k \left( \frac kp \right) $$ where $\chi(k)=\left( \frac kp \right)$.
By Siegel's theorem, we have an ineffective (we know the existence of $C_1(\epsilon)>0$ but cannot write it explicitly) estimate $$ L(1,\chi)> C_1(\epsilon) p^{-\epsilon}. $$ Take $\epsilon=1/4$ and let $C=C_1(1/4)$. Then $$ \sum_{k=1}^{p-1} \frac1k \left( \frac kp \right)= L(1,\chi)-\sum_{k=p}^{\infty} \frac1k \left( \frac kp \right)> \frac C{p^{1/4}} - \frac{2\log p}{p^{1/2}}. $$ Therefore, for sufficiently large $p$ (meaning that there exists $p_0$ such that for all $p\geq p_0$), we have $$ \sum_{k=1}^{p-1} \frac1k \left( \frac kp \right)>0. $$