For a positive integer $n\ge 3$, let $A_n=\{\ln 3, \ln 4,\dots,\ln n\}$. Does there exist $N$ such that for all $n>N$, the set $A_n$ can be partitioned into two sets so that their sums differ by no more than $1$?
I think for odd $n$ it might be possible to partition the set as $\{\ln 3,\ln 5,\dots,\ln n\}$ and $\{\ln 4,\ln 6,\dots,\ln (n-1)\}$. The difference of sums is $\ln\left(\frac{3\cdot 5\cdot\dots n}{4\cdot 6\cdot\dots(n-1)}\right)$. The ratio within $\ln$ shouldn't be too large (we require it to be no more than $e$).
$$ \ln\frac{(k+1)(k+2)}{k(k+3)}=\ln\frac{k^2+3k+2}{k^2+3k}=\ln\left(1+\frac2{k^2+3k}\right)\lt\frac2{k^2+3k}\;. $$
Thus
\begin{align} \sum_{k=0}^\infty\left(-\ln(j+4k)+\ln(j+4k+1)+\ln(j+4k+2)-\ln(j+4k+3)\right) &\lt\sum_{k=0}^\infty\frac2{(j+4k)^2+3(j+4k)} \\ &\le\sum_{k=0}^\infty\frac2{(1+4k)^2+3(1+4k)} \\ &= \frac\pi{12}+\frac{\ln2}2 \\ &\lesssim0.61\;. \end{align}
So we only need one set for each non-zero residue modulo $4$ that has difference less than $0.39$, e.g.:
$$ \ln3+\ln4+\ln5-\ln6-\ln7=\ln\frac{10}7\lesssim0.36\;,\\ \ln4+\ln5+\ln7-\ln3-\ln6-\ln8=\ln\frac{35}{36}\gtrsim-0.03\;,\\ \ln3+\ln4+\ln5+\ln7-\ln6-\ln8-\ln9=\ln\frac{35}{36}\gtrsim-0.03\;. $$