Answer: $2$. Example: $n=p^8q^8$ and there is $p^{i-1}q^{j-1}$ in $i.$ column $j.$ row. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1&p&p^2&p^3&p^4&p^5&p^6&p^7&p^8 \\\hline q&pq&p^2q&p^3q&p^4q&p^5q&p^6q&p^7q&p^8q \\\hline q^2&pq^2&p^2q^2&.&.&.&.&.&. \\\hline q^3&pq^3&.&p^3q^3&.&.&.&.&. \\\hline q^4&pq^4&.&.&p^4q^4&.&.&.&. \\\hline q^5&pq^5&.&.&.&p^5q^5&.&.&. \\\hline q^6&pq^6&.&.&.&.&p^6q^6&.&. \\\hline q^7&pq^7&.&.&.&.&.&p^7q^7&. \\\hline q^8&pq^8&.&.&.&.&.&.&p^8q^8 \\\hline \end{array} $$ Proof: If $p|n,$ then $v_p(n)\in \{2,8,80\}$. Hence $n$ has at most $4$ prime divisors. If $n$ has $4$ prime divisors, then call them $p,q,r,s$. $p^2q^2$ exists in some square which is in the same row or column with $16$ other numbers. The numbers which divide $p^2q^2$ and not equal to $p^2q^2$ are $\{1,p,p^2,q,q^2,pq,p^2q,pq^2\}$. The numbers which can be divided by $p^2q^2$ and not equal to $p^2q^2$ are $\{p^2q^2r,p^2q^2r^2,p^2q^2s,p^2q^2s^2,p^2q^2rs,p^2q^2r^2s,p^2q^2rs^2,p^2q^2r^2s^2\}$. There exist $16$ numbers in these sets thus all numbers in these sets must be in the same row or column with $p^2q^2$. But any pair in the set $\{p^2,q^2,pq\}$ must be in different row and columns which gives a contradiction. If $n$ has $3$ prime divisors, call them $p,q,r$. WLOG $v_p(n)=8,v_q(n)=2,v_r(n)=2$. $p^8$ exists in some square. It's in the same row or column with $16$ other numbers. The divisors of $n$ which divide $p^8$ and different from it are $\{1,p,p^2,p^3,p^4,p^5,p^6,p^7\}$ and the divisors of $n$ which can be divided by $p^8$ and different from it are $\{p^8q,p^8q^2,p^8r,p^8r^2,p^8qr,p^8q^2r,p^8qr^2,p^8q^2r^2\}$. There are $16$ numbers in these sets thus all of them must be in the same row or column with $p^8$. But all pairs in the set $\{p^8q^2,p^8r^2,p^8qr\}$ must be in different rows and columns which gives a contradiction. So $n$ has at most $2$ prime divisors as desired.$\blacksquare$