jmc

Note that: $$\begin{array}{rl}1\cdot 1=1& \equiv 1(\mathrm{mod}13)\\ 2\cdot 7=14& \equiv 1(\mathrm{mod}13)\\ 3\cdot 9=27& \equiv 1(\mathrm{mod}13)\\ 4\cdot 10=40& \equiv 1(\mathrm{mod}13)\\ 5\cdot 8=40& \equiv 1(\mathrm{mod}13)\\ 6\cdot 11=66& \equiv 1(\mathrm{mod}13)\end{array}$$ Thus the modulo $13$ inverses of $1,2,3,4,5,6$ are $1,7,9,10,8,11,$ respectively. The only residue from $7$ through $12$ that is not the inverse of a residue from $1$ through $6$ is $\boxed{12}$ (which is its own inverse).

(Note that $m-1$ is always its own inverse modulo $m$, so we know without doing most of the work above that Dayna could not have erased $12$. The rest of the work only goes to confirm that $12$ is the $\text{only}$ residue she did not erase.)