Let $N$ be a positive integer that is divisible by $2010$ and that has exactly $2010$ positive divisors. Since $2010 = 2 \cdot 3 \cdot 5 \cdot 67$, also $N$ should be divisible by these four primes. Thus, $N = 2^a \cdot 3^b \cdot 5^c \cdot 67^d \cdot s$, where $a, b, c, d > 0$ and $s$ is not divisible by any of the primes $2, 3, 5, 67$. All the factors of $N$ can be expressed as $2^i \cdot 3^j \cdot 5^k \cdot 67^l \cdot t$, where $0 \le i \le a, 0 \le j \le b, 0 \le k \le c, 0 \le l \le d$, and $t$ is a factor of $s$. There are $a+1$ choices for $i$ (from $0$ to $a$) and similarly, there are $b+1, c+1$ and $d+1$ choices for $j, k$ and $l$, respectively. Therefore, $N$ has $\delta(N) = (a+1)(b+1)(c+1)(d+1)\delta(s)$ different factors, where $\delta(x)$ stands for the number of factors of $x$. We require $\delta(N) = 2010$. As $a+1 > 1, b+1 > 1, c+1 > 1$, and $d+1 > 1$, we see that each of these numbers is divisible by some prime numbers and the number $\delta(N) = (a+1)(b+1)(c+1)(d+1)\delta(s)$ can thus be expressed as a product of at least four prime numbers. But as $2010$ itself is a product of exactly four prime numbers, we conclude that $a+1, b+1, c+1$, and $d+1$ are exactly those primes $2, 3, 5$, and $67$, in some order, and $\delta(s) = 1$. From the latter condition we see that $s=1$ because any numbers bigger than $1$ has more than one factor. So for $N$ to satisfy the conditions, $N$ must be expressible as $2^a \cdot 3^b \cdot 5^c \cdot 67^d$, where $a, b, c, d$ are the numbers $1, 2, 4$, and $66$ in some order. Thus there are $4! = 24$ numbers satisfying the conditions.