The answer is no. First, for every $1 \leq i \leq 8$ we define $N_i$ to be the set of numbers assigned to the vertices adjacent to the vertex with number $i$. By the problem's assumption, the sum of elements of $N_i$ is divisible by $i$. Since the sum of every three numbers less than $8$ is at most $7+6+5=18$, the sum of elements of $N_8$ is either $8$ or $16$. By checking different cases, it is easy to see $N_8$ equals to either $\{3, 6, 7\}$, $\{4, 5, 7\}$, $\{5, 2, 1\}$, or $\{4, 3, 1\}$. We will check these different cases.

* If $N_8 = \{3, 6, 7\}$, let $N_6 = \{x, y, 8\}$. Evidently $N_8 \cap N_6 = \emptyset$. Now, by assumption, $6 \mid x+y+8$, and $11 = 1+2+8 \leq x+y+8 \leq 17 = 4+5+8$. These imply $x+y = 4$ and so $\{x, y\} = \{1, 3\}$. This contradicts $N_8 \cap N_6 = \emptyset$.

* The case $N_8 = \{4, 5, 7\}$ is similar to the previous case. We only need to consider $N_7$ instead of $N_6$.

* If $N_8 = \{1, 3, 4\}$, denote by $x, y, z, t$ the numbers of other vertices as in the figure below. We have $\{x, y, z, t\} = \{2, 5, 6, 7\}$. It follows from $t \mid x+y+z$ that $t$ is a divisor of $t+x+y+z = 20$ so $t = 5$. On the other hand, $y \mid 1+3+t$. Therefore $y$ must be either $1$, $3$ which is impossible.

* If $N_8 = \{5, 2, 1\}$, denote the numbers of other vertices by $x, y, z, t$ as in the figure below. In this case $\{x, y, z, t\} = \{3, 4, 6, 7\}$. By a similar argument to the previous case, we have $t = 4$. On the other hand, $x \mid 1+5+t$, so $x$ is a divisor of $10$, again impossible.

Therefore, in each case, the numbers cannot be assigned to the vertices of the cube and the solution is complete. ■