For $x=(x\_1,...,x\_n)\in \mathbb{R}^n$, let $Q\_x=(x\_1,x\_1+1)\times ...\times (x\_n,x\_n+1)$ - the open cube having $x$ in its "bottom left" corner. It seems, I can prove (see a draft [here](https://drive.google.com/open?id=1CDCUqG0aJhmrw35NazFTlzTR0fpccFcL)) the following 


**Theorem.** Let $K\subset\mathbb{R}^n$. There is $v\in \mathbb{R}^n$ so that 
for any $m\in\mathbb{Z}^n$, either $v+K\cap Q\_m\ne\varnothing$, or $v+K\cap \overline{Q\_m}=\varnothing$. In fact, such $v$'s for a co-meager set in $\mathbb{R}^n$.


This means that we can shift the integer grid so that there is no cube that intersects $K$ only by the boundary (of the cube).


My proof is not very clever, but rather lengthy, and I am not sure I like the way it is written.



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**Edit:** Based on Mathieu Baillif's comment it was possible to give a rather simple proof of a similar fact in the context of general Baire topological group. I am however still looking for a reference.



Reference request: placing a set with respect to the integer grid
Based on Mathieu Baillif's comment we can actually prove a much more abstract 


**Theorem.** Let $G$ be a Cech-complete topological group, let $U\_n$ be a sequence of non-empty open sets in $G$, and let $K\subset G$. Then there is $g\in G$ such that for every $n$ either $gK\cap U\_n\ne \varnothing$ or $gK\cap \overline{U\_n}= \varnothing$. Moreover, such $g$'s form a co-meager set.


Of course, in the context of the original question $G$ is the Euclidean space, and $U\_n$ is the sequence of the cubes formed by the integer grid.


**Proof.** Observe that $A,B\subset G$ we have that $\{g\in G: gA\cap B\ne\varnothing\}=BA^{-1}$. Indeed, $gA\cap B\ne\varnothing$ iff there are $a\in A$ and $b\in B$ such that $ga=b$ or $g=ba^{-1}$.


Let $H\_n=\{g\in G,~ gK\cap U\_n= \varnothing,~ gK\cap \overline{U\_n}\ne \varnothing\}$. Then, $H\_n=\overline{U\_n}A^{-1}\backslash U\_nA^{-1}$. Since $U\_n$ is open, so is $U\_nA^{-1}$ as a union of open sets. Hence, $\overline{H\_n}\subset\overline{\overline{U\_n}A^{-1}}\backslash U\_nA^{-1}$.


Since the product is continuous from $G\times G$ into $G$ it follows that $$\overline{U\_nA^{-1}}\subset \overline{\overline{U\_n}A^{-1}}\subset \overline{\overline{U\_n}\cdot\overline{A^{-1}}}=\overline{U\_n\cdot A^{-1}},$$ from where $\overline{H\_n}\subset\overline{U\_nA^{-1}}\backslash U\_nA^{-1}=\partial (U\_nA^{-1})$. A boundary of an open set is nowhere dense, and so $H\_{n}$ is nowhere dense.


The set $G\backslash \bigcup H\_n$ is co-meager (and so non-empty). If $g\in G\backslash \bigcup H\_n$, then for every $n$ $g\notin H\_n$, from where either $gK\cap U\_n\ne \varnothing$ or $gK\cap \overline{U\_n}= \varnothing$. $\square$


