$1+\lceil \frac n2 \rceil$

Clearly it suffices to bound for $n$ odd. Global + Equality: Call a square X if it is greater than all numbers to its left, and Y if it is greater than all numbers above it. Note the first column is all X and first row is all Y. This means that in the $(n-1)\cdot (n-1)$ subgrid not containing the first row and first column of the original grid. In this subgrid, each row contains at most $\frac{n-1}{2}$ X's and each column contains at most $\frac{n-1}{2}$ Y's. Summing, $(n-1)^2 \le \# X + \# Y \le \frac{n-1}{2} (n-1+n-1)=(n-1)^2$ This means equality holds, each column and row has exactly $\frac{n-1}{2}$ X's and Y's, and each square can only be one of X,Y. Extremal (the hard part): We use the property that each square can be one of $X,Y$. Consider the grid with the largest value in this $(n-1) \cdot (n-1)$ subgrid, call $x_{k,j}$. If it is marked X, then consider its column. If there is a Y below it, it is not maximal. Otherwise, there is a Y above it (and between this X and the Y there are only X's), say at $x_{k,y}$. Then we can induct on $l$ to show for all $l>j, x_{k,l}<x_{k,j}$, or $x_{k,l}=max\{ x_{k,1}, \cdots, x_{k,l} \}$ contradicting $l$ being only $X$. Therefore, this grid cannot be the largest. Similarly, the largest number cannot have be marked Y, contradiction!

It clearly suffices to construct for $n$ even. Let $n=2k$. Divide the board into 4 $k\cdot k$ squares. For $i,j\ge 2$: if $i=j>k$ then mark them with both X and Y. If $i\le k$ and $j\le k$ or if $i>k$ and $j>k$ mark it with $X$ If $i>k$ and $j\le k$ or if $i\le k$ and $j>k$ or if $i=k+1$ and $j>k$ mark it with Y. So construct as follows: $x_{1,1}=10^{2k^2}.$ For $i\ge 2, x_{i,1}=i, x_{i,j}=i+jk^{-2}$. Set $x_{1,i}=k+i$ where $2\le i\le k$. Set $x_{i,j}=k^{-2}+ik^{-6}+jk^{-4}$ for $j\in \{k+1,\cdots,2k\}, i\in \{1,\cdots,k\}$. They are only marked Y's because $x_{i,j}<x_{i,k}$. Set $x_{j,1}=2^j$ for $j=k+1,\cdots, 2k$. Set $x_{j,t}=j+tk^{-2}$ for $t\in \{2,\cdots,k\}$ and $j\in \{k+1,\cdots,2k\}$. Set $x_{k+1,t}=10^{(k+1)^t}$ for $t\in \{k+1,\cdots,2k\}$ Set $x_{l,t}=2^l+t\epsilon$ for $l\ge k+2, t\ge k+1$.