Do you mean the maximum or minimum number of such squares that we can obtain?

I meant to find a square with min. number of rectangles with area <=1

Lemma. If rectangle $S$ has side lengths $1-x,y$ so that $(1-x)y>1$ then $y>1+x$. Consider general case for $n \times n$ square. We prove that there are at least $n$ rectangles with area not exceeding $1$. Let $V,H$ be set of all vertical, horizontal side lengths of small rectangles respectively then $|V|=|H|=n$. In order not to have at least $n$ rectangles whose areas $ \le 1$, there exists at least one $1-a \in V, 0<a<1$ and one $1-b \in H, 0<b<1$. WLOG, $1>a>b$. Case 1. If there is only one $1-b \in H$ (i.e. only one horizontal side length $<1$). Consider row of $n$ rectangles whose vertical length is $1-a$. Let $y_1,y_2, \ldots, y_{n-1},y_n \; (y_n=1-b)$ be horizontal side length of such rectangles then $\sum_{i=1}^ny_i=n$ and $y_i \ge 1$ for all $i= 1, \ldots, n-1$ according to the assumption. If there exists $y_{n-1}$ so $y_{n-1}>1+a$. Hence, $$(n-2) \min_{1 \le i \le n-2} \{y_i \} +1+a+1-b<n \implies \min_{1 \le i \le n-2} \{ y_i \} < 1- \frac{a-b}{n-2}<1.$$This contradicts to assumption that $y_i \ge 1$ for all $i=1, \ldots, n-1$. This follows all $y_i \le 1+a$ for $i=1, \ldots, n-1$, which deduces all $n$ rectangles have areas $\le 1$ according to Lemma. Case 2. From case 1, we find that each $V,H$ has at least two side lengths that are less than $1$. From $V \cup H$, arrange all side lengths $<1$ in order $0<1-a \le 1-a_1 \le 1-a_2\le \ldots <1$. WLOG, $1-a \in V$. Let $n-m$ be number of horizontal side lengths $> 1+a$ then $m(1-a)+(n-m)(1+a) < \sum_{\ell \in H} \ell =n$ which follows $m>n/2$, i.e. there are at least $n/2$ horizontal side lengths $\le 1+a$. By considering row of rectangles whose vertical side length is $1-a$, then the row consists at least $n/2$ rectangles areas $\le (1-a)(1+a)<1$. If there is another vertical side length of $1-a$ then we are done. If not, all the rest $n-1$ vertical side lengths have length $\ge 1-a_1$. If there exists a horizontal side length $\le 1-a_1$ then with similar argument, we can find columns of rectangle whose horizontal side length is $1-a_1$ have at least $n/2$ rectangles whose areas $\le 1$ and can conclude. Next consider case for all $n-1$ horizontal side lengths (except for $1-a$) are $\ge 1-a_2$ and we keep going. Since we only have limited number of side lengths $<1$ so this process will end showing that there exists at least $n$ rectangles whose areas $\le 1$. Construction for $n$ rectangles can be followed from idea of case $1$: Pick one horizontal side length $\varepsilon$ very small, pick one vertical side length to be $0.99$. Others horizontal side lengths are the same and equal $\frac{n-\varepsilon}{n-1}$ which can give $\frac{n-\varepsilon}{n-1} \cdot 0.99>1$. Other vertical side lengths are the same and equal $\frac{n-0.99}{n-1}>1$. With this, only $n$ rectangles whose horizontal side length is $\varepsilon$ have area $\le 1$.

Quote: Lemma. If rectangle $S$ has side lengths $1-x,y$ so that $(1-x)y>1$ then $y>1+x$. Consider general case for $n \times n$ square. We prove that there are at least $n$ rectangles with area not exceeding $1$. Let $V,H$ be set of all vertical, horizontal side lengths of small rectangles respectively then $|V|=|H|=n$. In order not to have at least $n$ rectangles whose areas $ \le 1$, there exists at least one $1-a \in V, 0<a<1$ and one $1-b \in H, 0<b<1$. WLOG, $1>a>b$. Case 1. If there is only one $1-b \in H$ (i.e. only one horizontal side length $<1$). Consider row of $n$ rectangles whose vertical length is $1-a$. Let $y_1,y_2, \ldots, y_{n-1},y_n \; (y_n=1-b)$ be horizontal side length of such rectangles then $\sum_{i=1}^ny_i=n$ and $y_i \ge 1$ for all $i= 1, \ldots, n-1$ according to the assumption. If there exists $y_{n-1}$ so $y_{n-1}>1+a$. Hence,$$(n-2) \min_{1 \le i \le n-2} \{y_i \} +1+a+1-b<n \implies \min_{1 \le i \le n-2} \{ y_i \} < 1- \frac{a-b}{n-2}<1.$$This contradicts to assumption that $y_i \ge 1$ for all $i=1, \ldots, n-1$. This follows all $y_i \le 1+a$ for $i=1, \ldots, n-1$, which deduces all $n$ rectangles have areas $\le 1$ according to Lemma. Case 2. From case 1, we find that each $V,H$ has at least two side lengths that are less than $1$. From $V \cup H$, arrange all side lengths $<1$ in order $0<1-a \le 1-a_1 \le 1-a_2\le \ldots <1$. WLOG, $1-a \in V$. Let $n-m$ be number of horizontal side lengths $> 1+a$ then $m(1-a)+(n-m)(1+a) < \sum_{\ell \in H} \ell =n$ which follows $m>n/2$, i.e. there are at least $n/2$ horizontal side lengths $\le 1+a$. By considering row of rectangles whose vertical side length is $1-a$, then the row consists at least $n/2$ rectangles areas $\le (1-a)(1+a)<1$. If there is another vertical side length of $1-a$ then we are done. If not, all the rest $n-1$ vertical side lengths have length $\ge 1-a_1$. If there exists a horizontal side length $\le 1-a_1$ then with similar argument, we can find columns of rectangle whose horizontal side length is $1-a_1$ have at least $n/2$ rectangles whose areas $\le 1$ and can conclude. Next consider case for all $n-1$ horizontal side lengths (except for $1-a$) are $\ge 1-a_2$ and we keep going. Since we only have limited number of side lengths $<1$ so this process will end showing that there exists at least $n$ rectangles whose areas $\le 1$. Construction for $n$ rectangles can be followed from idea of case $1$: Pick one horizontal side length $\varepsilon$ very small, pick one vertical side length to be $0.99$. Others horizontal side lengths are the same and equal $\frac{n-\varepsilon}{n-1}$ which can give $\frac{n-\varepsilon}{n-1} \cdot 0.99>1$. Other vertical side lengths are the same and equal $\frac{n-0.99}{n-1}>1$. With this, only $n$ rectangles whose horizontal side length is $\varepsilon$ have area $\le 1$. So what's the answer?