, 2007

Let there be $k$ kinds of numbers on the cards. Let $a_1 < a_2 < \dots < a_k$ be these numbers and each number is written on $p_1, p_2, \dots, p_k$ cards respectively. We will show $(p_1+1)(p_2+1)\dots(p_n+1) = 2008$, $a_1 = 1$ and $a_i = (p_1+1)(p_2+1)\dots(p_{i-1}+1)$ ($1 < i \le k$).

Since it is able to choose some cards which sum up to $1$, $a_1 = 1$. Only the numbers less than or equal to $p_1$ can be made with $a_1$s, so if $k \ge 2$ we get $a_2 = p_1+1$. Then only the numbers less than or equal to $a_1p_1 + a_2p_2 = (p_1+1)(p_2+1) - 1$ can be made with $a_1$s and $a_2$s, so if $k \ge 3$ we get $a_3 = (p_1+1)(p_2+1)$. Continuing this argument we get $a_i = (p_1+1)(p_2+1)\cdots(p_{i-1}+1)$ for $1 < i \le k$. Since $a_1p_1 + a_2p_2 + \cdots + a_kp_k = 2007$, $(p_1+1)(p_2+1)\cdots(p_n+1) = 2007+1 = 2008$.

On the other hand, given some positive integers $p_1, p_2, \dots, p_k$ with $(p_1+1)(p_2+1)\cdots(p_n+1) = 2008$, letting $a_1 = 1$ and $a_i = (p_1+1)(p_2+1)\cdots(p_{i-1}+1)$ ($1 < i \le k$) satisfies the condition.

So all we have to do is to count $(p_1, p_2, \dots, p_k)$ which $(p_1+1)(p_2+1)\cdots(p_n+1) = 2008$ holds. Since $2008 = 2^3 \times 251$ and $2$ and $251$ are primes, $(p_1+1) \times (p_2+1) \times \cdots \times (p_n+1)$ must be $2008, 2 \times 1004, 4 \times 502, 8 \times 251, 2 \times 2 \times 502, 2 \times 4 \times 251, 2 \times 2 \times 2 \times 251$ or one of their permutations. There are $1, 2, 2, 2, 3, 6, 4$ possible permutations respectively, so the answer is $1+2+2+2+3+6+4=20$.