Let a $(2k-1) \times (2k-1)$ square board be covered as in the statement with $x$ squares $2 \times 2$ and $y$ shapes L. Denote by $(i, j)$ the cell in row $i$, column $j$, and color black all cells $(i, j)$ with both $i$ and $j$ odd. Thus $k^2$ black cells are obtained. Observe that wherever a $2 \times 2$ square is placed, it covers exactly one black cell; and wherever an L-shape is placed, it covers at most one black cell. To have the whole board covered it is necessary that the total number of figures be at least $k^2$, i.e. $x+y \ge k^2$.

All figures cover $4x+3y$ cells, which equals $(2k-1)^2$, the total number of cells on the board. On the other hand $x \ge k^2-y$ implies $4x+3y \ge 4(k^2-y)+3y = 4k^2-y$. Hence $4k^2-y \le (2k-1)^2$, yielding $y \ge 4k-1$. In summary each admissible covering has at least $4k-1$ shapes L and at most $k^2-4k+1$ squares $2 \times 2$.

A $13 \times 13$ board corresponds to the case $k=7$, so the number of L-shapes is at least $4 \cdot 7 - 1 = 27$; the number of $2 \times 2$ squares is at most $7^2 - 4 \cdot 7 + 1 = 22$. The example in the figure shows a covering with 22 squares $2 \times 2$ and 27 shapes L. Hence the minimum number of L-shapes is 27.