8. We will call two numbers almost equal to each other if they are equal to each other or differ from each other by no more than one. A rectangular grid with sides of natural numbers a and b is such that it is impossible to cut out a rectangle along the grid lines, the area of which is almost equal to half the area of the original rectangle. What is the smallest value that the number $|a-b|$ can take? (E. Molchanov, S. Berlov)
Answer: 4. Solution: If one of the sides of the rectangle is even, a rectangle of half the area can be cut from it along the midline. Therefore, it is sufficient to consider the case where both sides are odd. In this case, $|b-a|$ is even. If $|b-a|=0$, then $a=b=2n+1$, half the area of the rectangle is $2n^2+2n+0.5$, and we can cut from it a rectangle of almost this area, $2n \times (n+1)$, with an area of $2n(n+1)$, and when $a=1$ - the square $1 \times 1$ with an area almost equal to $1/2$. If $|b-a|=2$, then $a=2n+1$, $b=2n-1$, half the area of the rectangle is $2n^2-0.5$, and we can cut from it a rectangle $2n \times n$ with an area almost equal to this number, $2n^2$. Thus, $|a-b| \geq 4$. Assuming $a=2n+3, b=2n-1$, we get a rectangle whose half area is $2n^2+2n-1.5$. Almost equal to it are the integer areas $2n^2+2n-1$ and $2n^2+2n-2$. For $n=3$, $2n^2+2n-1=23$ and $2n^2+2n-2=22=2 \cdot 11$. The prime numbers 23 and 11 are greater than the sides of the resulting rectangle $9 \times 5$ when $n=3$, so it is impossible to cut from it a rectangle along the grid lines with an area almost equal to half its area. Thus, the minimum possible value of $|a-b|$ is 4.