Solution:

Let $N = (2k+1) + (2k+3) + \cdots + (2k+2n-1)$ where $n$ and $k$ are integers, $n \geq 2$, $k \geq 0$. Then
$$N = [(2k+1) + (2k+2n-1)] \frac{n}{2} = (2k+n)n,$$
which is a product of two integers with the same parity, since adding the even number $2k$ to the integer $n$ does not change its parity.

1. Since $2005 = 401 \cdot 5$, taking $n = 5$ gives $2k+5 = 401$, i.e. $2k+1 = 397$. Checking, we see that $397 + 399 + 401 + 403 + 405 = 401 \cdot 5 = 2005$, so $2005$ can be represented as the sum of at least two consecutive odd positive integers.

2. Since $2006 = 2 \cdot 1003 = 2 \cdot 17 \cdot 59$ is divisible by $2$ but not by $4$, every pair of integers with a product of $2006$ consists of two integers with different parity (i.e. one odd and one even). Therefore $2006$ cannot be represented as the sum of at least two consecutive odd positive integers.