Motivation (can skip!). (*) $\sum\log n \approx n\log n-n,$ and $$\sum\log n = \sum_{p_1\leq n} \log p_1+\sum_{p_2\leq n} \log p_2+...+\sum_{p_m\leq n} \log p_m$$
in which $p_k$ are numbers comprised of $k$ primes including repetitions. (*) is not an asymptotic equality but almost everything that is known about $\sum \log p_k$ is due to the prime number theorem.
The following is maybe obvious but I can't prove it yet. I'd be interested in seeing a proof or a hint (thanks). In words: the (sums of) odd summands are $\sim$ the (sums of) even summands.
Let $n = 2^{m},~ m = 2 r.$ Let $p_k(i)$ be the i$^{th}$ k-prime.
$$\sum_{k=1}^r\sum_{p_{2k}(i)\leq n}\log p_{2k}(i)\sim \sum_{k=1}^r\sum_{p_{2k-1}(i)\leq n}\log p_{2k-1}(i)\sim (1/2)(n\log n-n) $$
For $m = 18,$
$\sum\sum_{2k}/(2^{18}\log 2^{18}-2^{18})\approx 0.49944$ and $\sum\sum_{2k-1}/(2^{18}\log 2^{18}-2^{18})\approx 0.50056$
This is equivalent to showing that $\sum_{k\le n} \lambda(k) \log k = o(\sum_{k\le n} \log k) = o(n \log n)$, where $\lambda(n) := (-1)^{\Omega(n)}$ is the Liouville lambda function. The Dirichlet series for $\lambda(n)$ is $\sum_{n\in \mathbb N} \lambda(n) n^{-s} = \zeta(2s)/\zeta(s)$, so one should be able to use the zero-free region to establish bounds on the summatory function $L(x):=\sum_{n\le x} \lambda(n)$. In particular, although I couldn't find a reference, I'd expect it's known that
$$L(x) = O(x e^{-c\sqrt{\log x}}).$$
for some constant $c>0$, and thus $|L(x)| \le Cx(\log x)^{-A}$ for each $A\ge 0$. Combining this with partial summation gives a similar result for $\sum_{k\le n} \lambda(k) \log k$, so it is indeed asymptotically small compared to $n \log n$.