Let $G$ be a group of order $n$, on elements $x_1, ..., x_n$.
Let's draw the Cayley table of the group, that is, table $A$ with $a_{i,j}$ = [number of $x_i \circ x_j$ in the list $x_1, ..., x_n$] After that, if we erase $K$ entries of the table, for some $K$ it's always possible to determine the deleted entries, and for some $K$ it's not possible since there exist some group $G'$ (maybe, isomorphic to $G$ and maybe not) whose Cayley table is different from $A$, but only in some of the erased cells.
The question is, whether it's always possible to determine original table (for all sufficiently large $n$) if $K=2n$.
Here's what I know so far:
if $K=4n-4$, the table can't be reconstructed when we delete 2 rows and 2 columns, corresponding to identity and some other element $x$ (so, after that we won't be able to determine the difference between them, and different tables for the same group count as different)
if $K=n$, it's possible for $n \neq 4$, because
if some column contains only one empty cell, we know what was written in that cell, since all elements in any column must be different
after that, you have at least $\lceil \frac{n}{2} \rceil$ columns in which you know every single entry
Subset of $G$, corresponding to such columns, form a subgroup of $G$, since if you know $x\circ y$ for any $x$, and $x\circ z$ for any $x$ in $G$, then you know $x \circ y \circ z$ for any $x$ in $G$, and $y^{-1}=y^{n-1}$. Let's denote that subgroup $H$.
So if you know more than $\frac{n}{2}$ columns, you can reconstruct the table, and the only case when you can't is when $n$ is even and in exactly $\frac{n}{2}$ columns exactly two entries are unknown.
The set of elements corresponding to not-yet-known columns is a coset of $H$ ($H$ is subgroup of index 2). That is, its elements are of the form $gh$, $h \in H$, where $g$ is fixed. To know what is $gh_i \circ gh_j$, it's sufficient to know which element equals $gh_i g$ (since we know how to multiply by elements of $H$). And to determine $gh_ig$ we look at cells corresponding to $gh_i \circ gh_j$ (they are in the same column) For all $j$ (except maybe two) we know the value of $g h_i g h_j$ and so, if $n>4$, we definitely know what is $gh_i g$.
The same for $h_i \circ gh_j$
So, I'm asking about $K=2n$. Those tricks with subgroup of index two won't work, but I don't think that there are counterexamples for arbitrarily large $n$.
Sorry for my English and unnecessary longishness of description. Feel free to edit the question.