Japan 2007

Let there be $k$ kinds of numbers on the cards. Let $a_1<a_2<\cdots <a_k$ 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 $a_1{p}_1+a_2{p}_2=(p_1+1)(p_2+1)-1$ 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 $a_1{p}_1+a_2{p}_2+\cdots +a_k{p}_k=2007$, $(p_1+1)(p_2+1)\cdots (p_n+1)=2007+1=2008$.

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 $1+2+2+2+3+6+4=20$.