Is there an $n \times n$ multiplication table such that if you form a border of width $k$ ("the frame") and sum its elements, the total will equal the sum of the remaining elements ("the picture")?
The diagram on the left is intended as a general representation. On the right, I created a $5 \times 5$ multiplication table (entries omitted) and summed them in their corresponding color. The width of the border was $k = 1$, and I found $\sum$frame $= 144$ but $\sum$picture $= 81$. So, in this case, the sums were not equal.
(Yes, there is a quick parity argument to show the sums here aren't equal, but I wanted to carry out the computations explicitly in an example to ensure that my question is understood.)
If the answer is no such table exists: how do you prove this?
If the answer is yes: what is the minimal $n$ (and its minimal $k$) for which this is possible?
I believe there is no such table.
Consider an $n \times n$ table with a border of $k$.
The sum of all entries in the table is $$ \left(\frac{1}{2}n(n+1) \right) ^2 $$ while the sum of all entries in the "picture" is $$ \sum_{i=k+1}^{n-k} \sum_{j=k+1}^{n-k} ij = \left( \sum_{i=k+1}^{n-k} i \right)^2 = (n+1)^2 \left(\frac{1}{2}n-k \right)^2$$
Thus, we seek a pair $n,k$ with $$ \frac{1}{2}\left(\frac{1}{2}n(n+1) \right) ^2 = (n+1)^2 \left(\frac{1}{2}n-k \right)^2.$$ This last equation simplifies to $$0=n^2-8kn+8k^2$$ from which we may conclude that $$n=(4 \pm \sqrt{8}) k.$$ Since $k$ and $n$ must be integers, we may conclude that there is no such pair.