bump, any solutions?

The answer is [n/2]-1.

Consider a row doesn't contains the original square and it's not the border,if the Hamilton road turns left or right when reach it,then one of the 2 rows next to it has a block,if the road always go straight when reach it, then consider the column we can know that at least n-2 columns has blocks,other wise consider pairs （U，U+2),less then one pair satisfies both the Uth and U+2 th row don't have blocks,so there is at least [n/2]-1 blocks. And the construction is for 4|n-1,we set blocks at the left of the 3,7,...row and right of 6,10,.....row,and begin with the second square at the third row,for the other n bigger than 5,we set blocks at the left of the 5,9,13,....row and right of 4,8,.....row and begin with the left of the second row.

The answer is $\left\lfloor\frac{n}{2}\right\rfloor-1$. But the construction is too troublesome! My answer with full of pics.

Solved with rama1728. We claim that the answer is $\lfloor{\frac{n}{2}\rfloor} -1$. Say we have a working configuration with $m$ blocks on it. One can notice that the robot can slide along a row if and only if there is a block or an edge above or below it. Thus, placing a block 'unlocks' two rows. From this we can note that, we require, \[2m +3 \geq n \implies m \geq \lfloor{\frac{n}{2}\rfloor} -1\]Now, we shall give a construction to show that this number of blocks is sufficient. Note that, when $n \equiv 1 \pmod{4}$, we can simply place blocks on rows $3,7,11,\dots$ in the left-most column and rows $6,10,\dots$ of the right-most columns and simply start from the third row. For all other $n$, simply place blocks on the left-most column on rows $5,9,13,\dots$ and on the right-most column for rows $4,8,12,\dots$ and begin on the second row. Thus, indeed our answer is correct and we are done.