We claim that this is only possible if $k$ has a units digit of 1. Clearly, it is true in these cases. Additionally, $k^{k}$ cannot have a units digit of 1 when $k$ has a units digit of $2,4,5,6$, or 8. If $k$ has a units digit of 3 or 7, then $k^{k}$ has a units digit of 1 if and only if $4 \mid k$, a contradiction. Similarly, if $k$ has a units digit of 9, then $k^{k}$ has a units digit of 1 if and only if $2 \mid k$, also a contradiction. Since there are 202 integers between 1 and 2013, inclusive, with a units digit of 1, there are 202 such $k$ which fulfill our criterion.