First, we note that $2010=10(201)=2(5)(3)(67)$ and so $2010^{2}=2^{2} 3^{2} 5^{2} 67^{2}$. Consider $N$ consecutive four-digit positive integers. For the product of these $N$ integers to be divisible by $2010^{2}$, it must be the case that two different integers are divisible by 67 (which would mean that there are at least 68 integers in the list) or one of the integers is divisible by $67^{2}$. Since we want to minimize $N$ (and indeed because none of the answer choices is at least 68), we look for a list of integers in which one is divisible by $67^{2}=4489$. Since the integers must all be four-digit integers, then the only multiples of 4489 the we must consider are 4489 and 8978. First, we consider a list of $N$ consecutive integers including 4489. Since the product of these integers must have 2 factors of 5 and no single integer within 10 of 4489 has a factor of 25 , then the list must include two integers that are multiples of 5 . To minimize the number of integers in the list, we try to include 4485 and 4490. Thus our candidate list is $4485,4486,4487,4488,4489,4490$. The product of these integers includes 2 factors of 67 (in 4489), 2 factors of 5 (in 4485 and 4490), 2 factors of 2 (in 4486 and 4488), and 2 factors of 3 (since each of 4485 and 4488 is divisible by 3). Thus, the product of these 6 integers is divisible by $2010^{2}$. Therefore, the shortest possible list including 4489 has length 6. Next, we consider a list of $N$ consecutive integers including 8978. Here, there is a nearby integer containing 2 factors of 5, namely 8975. So we start with the list $8975,8976,8977,8978$ and check to see if it has the required property. The product of these integers includes 2 factors of 67 (in 8978), 2 factors of 5 (in 8975), and 2 factors of 2 (in 8976). However, the only integer in this list divisible by 3 is 8976 , which has only 1 factor of 3 . To include a second factor of 3 , we must include a second multiple of 3 in the list. Thus, we extend the list by one number to 8979 . Therefore, the product of the numbers in the list $8975,8976,8977,8978,8979$ is a multiple of $2010^{2}$. The length of this list is 5 . Thus, the smallest possible value of $N$ is 5 .