TITLE: Least possible number of squares with odd side length
QUESTION [5 upvotes]: An $n\times(n+3)$ rectangular grid ($n>10$) is cut into some squares, with all cuts being along the grid lines. What is the least possible number of squares with odd side length?
[Source: Russian competition problem]

REPLY [2 votes]: Denote by $f(n)$ the minimal number of odd-sized squares in a decomposition
of a $n\times (n+3)$ square. Then (as was guessed by @Willemien in his answer)
we have $f(n)=g(n)$ where $g(n)=4$ when $n$ is of the form $4q$ or $4q+1$, and
$g(n)=2$ when $n$ is of the form $4q+2$ or $4q+3$.
First, if $r$ is the number of odd-sized squares in any decomposition
of a $n\times (n+3)$ square, counting the unit squares and working modulo
$4$, we see that $n(n+3)\equiv r \ ({\textsf{mod}} \ 4)$. Also, $r$
cannot be zero because one of $n$ or $n+3$ is odd. It follows that
$r\geq g(n)$ for any decomposition, and hence $f(n)\geq g(n)$.
The converse inequality $f(n)\leq g(n)$ has already been shown in
@Willemien's answer. Below is a picture summarizing his argument (with
even-sized rectangles in green and odd-sized squares in red ):