is easy to find one exemple for n = 13. to prove that for $n\geq 15$ is impossible, divide the table in $E(\frac{n}{2})$ dominoes. every dominoe will have at least one coin! so, for $n\geq 15$ is impossible, because we have at least 112 coins! for n = 14, we will use other fact. in any 2x2 table, the sum of the number of coins in your entries is even (is easy to see). so, we can divide one 14x14 table in 49 2x2 tables, and the sum of coins in our table will be even! (impossible, because 111 is odd)

For $n=13$, an example is as follows. Put $1,0,1,\dots,1$ (total$=7$) coins to odd-indexed rows, and $0,1,0,\dots,0$ (total$=6$) to even-indexed rows. The total number of coins is $85$. Now, note that, the zeroes are all surrounded by $1$'s, hence, making them $2$ do not change anything. Simply select $13$ cells inside with $0$'s in it, and update their status to $2$ coins.

n=14.parity answer n=13