, 2019

Solution:

By adding up all the equations on each vertex, we get $2019 S = k S$ where $S$ is the sum of all entries, so $k = 2019$ unless $S = 0$. In the latter case, by adding up all the equations on a half of the cube, we get $2018 S - S = k S$ where $S$ is the sum of all entries on that half of the cube, so $k = 2017$ unless $S = 0$. In the latter case (the sum of all entries of any half is zero), by adding up all the equations on a half of the half-cube, we get $2017 S - 2 S = k S$, so $k = 2015$ unless $S = 0$. We continue this chain of casework until we get that the sum of every two vertices connected by unit segments is zero, in which case we have $k = -2019$. This means that $k$ can take any odd value between $-2019$ and $2019$ inclusive, so the sum of absolute values is $2 \cdot 1010^{2} = 2040200$.

To achieve these values, suppose that the vertices of the hypercube are $\{0,1\}^{2019}$ and that the label of $\left(x_{1}, x_{2}, \ldots, x_{2019}\right)$ is $a_{1}^{x_{1}} a_{2}^{x_{2}} \ldots a_{2019}^{x_{2019}}$ for constants $a_{1}, a_{2}, \ldots, a_{2019} \in \{-1,1\}$, then it is not difficult to see that this labeling is $\left(a_{1} + a_{2} + \cdots + a_{2019}\right)$-harmonic for any choice of $a_{i}$'s, so this can achieve all odd values between $-2019$ and $2019$ inclusive.