The answer is $\left\lfloor \dfrac{(n+1)(m+1)}{2} \right\rfloor - 1$.

For a shelf $R$, let $\tilde{R}$ be the rectangular area by extending $R$ by 1 row to the top and 1 column to the left. Let us also extend the initial chessboard by 1 row to the top and 1 column to the left. We assign $-\infty$ to each of those new unit squares so that the number of shelves is preserved. For a shelf $R$, let us call the integer $h$ satisfying two conditions the *height* of $R$.

Claim 1: If two shelves $R_1$ and $R_2$ do not intersect, then $\tilde{R}_1 \cap \tilde{R}_2 = \emptyset$.

Proof of Claim 1: Let $h_1, h_2$ be the height of $R_1, R_2$ respectively. Without loss of generality, let us assume $h_1 \le h_2$. If $\tilde{R}_1 \cap \tilde{R}_2 \ne \emptyset$, then some unit square $x$ in $R_2$ is adjacent to $R_1$ and therefore the integer in $x$ is less than or equal to $h_1$. Since $x \in R_2$, the integer in $x$ is larger than $h_2$, contradictory to the assumption that $h_1 \le h_2$. This proves Claim 1.

Claim 2: If $R_1$ and $R_2$ are distinct maximal shelves smaller than the whole chessboard, then $R_1 \cap R_2 = \emptyset$.

Proof of Claim 2: Let $h_1, h_2$ be the height of $R_1, R_2$ respectively. Without loss of generality, let us assume $h_1 \le h_2$. Suppose that $R_1 \cap R_2 \ne \emptyset$. Suppose a unit square $x$ in $R_2$ shares at least one point with $R_1$. If $x$ does not belong to $R_1$, then the integer in $x$ should be at most $h_1$ but since $x \in R_2$, the integer in $x$ should be larger than $h_2$, contradictory to the assumption that $h_1 \le h_2$. So such $x$ belongs to $R_1$ and therefore $R_2 \subseteq R_1$. However this contradicts to the assumption that $R_2$ is maximal. So Claim 2 is proved.

Claim 3: The number of shelves in the $n \times m$ chessboard is less than or equal to $\dfrac{(n+1)(m+1)}{2} - 1$.

Proof of Claim 3: We proceed by induction on $n+m$. If $n+m=2$, then $n=m=1$ and the number of shelves is 1.

Now let us assume $n+m > 2$. Let $R_1, R_2, \dots, R_k$ be the maximal shelves strictly smaller than the whole $n \times m$ chessboard.

By Claims 1 and 2, $\tilde{R}_i \cap \tilde{R}_j = \emptyset$. By the induction hypothesis, the number of shelves contained in $R_i$ for each $i$ is at most $|\tilde{R}_i|/2 - 1$. Therefore the number of all shelves is at most
$$\sum_{i=1}^{k} \left( \left\lfloor \dfrac{|\tilde{R}_i|}{2} \right\rfloor - 1 \right) \le \dfrac{(n+1)(m+1)}{2} - k - 1.$$
So Claim 3 is proved if $k > 1$. We may now assume that $k=1$. Then it is enough to show that
$$\max \left( \dfrac{m(n+1)}{2} \cdot \dfrac{(m+1)n}{2} \right) - 1 + 1 \le \dfrac{(n+1)(m-1)}{2} - 1.$$
It is easy to see this because $n \ge 1$ is equivalent to the inequality that $\dfrac{m+n-1}{2} \le \dfrac{(n+1)(m+1)}{2} - 1$. This proves Claim 3.

Now it remains to show that there is an assignment of integers so that the number of shelves is exactly $\left\lceil \dfrac{(n+1)(m+1)}{2} \right\rceil - 1$. We proceed by induction on $n+m$. It is trivial if $\max(n,m) \le 2$.

Now by symmetry let us assume that $m > 2$. Let us write 1 in each unit square on the second row. For the first row, we write integers larger than 1 obtained by the induction hypothesis on a $1 \times n$ chessboard. For the remaining rows, we write integers larger than 1 obtained by the induction hypothesis on a $(m-2) \times n$ chessboard.

Now the number of shelves in this assignment is
$$1 + \left\lfloor \dfrac{2(n+1)}{2} - 1 \right\rfloor + \left\lceil \dfrac{(m-1)(n+1)}{2} - 1 \right\rceil$$
by the induction hypothesis. This completes the proof. $\Box$