Initially one have the number $0$ in each cell of the table $29 \times 29$. A moviment is when one choose a sub-table $5 \times 5$ and add $+1$ for every cell of this sub-table. Find the maximum value of $n$, where after $1000$ moviments, there are $4$ cells such that your centers are vertices of a square and the sum of this $4$ cells is at least $n$. Note: A square does not, necessarily, have your sides parallel with the sides of the table.