Let us show that number $11$ cannot be represented as a product of the odd parts of two consecutive positive integers. Assume the opposite: let $11 = x \cdot y$, where $x$ and $y$ are the odd parts of two consecutive positive integers. As $11$ is a prime number, either $x = 1$ and $y = 11$ or $x = 11$ and $y = 1$. As of the two consecutive integers one is always odd and the odd part of an odd number is the number itself, either $x$ or $y$ is one of the two consecutive integers. If it is $1$, then the other number can only be $2$, but the odd part of $2$ is not $11$. If it is $11$, then the other number can only be $10$ or $12$, but neither of those has odd part equal to $1$. In all cases we got a contradiction which proves the statement.

Let us show $11$ cannot be represented in the required way. If this representation existed, then due to the primality of $11$ the factors would have to be $1$ and $11$. Therefore one of the two consecutive integers has to be divisible by $11$. But this number cannot be $11$ itself, since neither $10$ nor $12$ has $1$ as its odd part. It also cannot be an odd multiple of $11$, because then its odd part would be the number itself rather than $11$. Finally, it cannot be an even multiple of $11$, since in such case the neighbouring numbers would be odd numbers greater than $1$, the odd parts of which are numbers themselves rather than $1$.