Replace $2003$ with $m$ and $10$ with $k$. So we want to show that given any number of societies with $k$ members such that any $m+1$ have $k+1$ with a common member, then there are $m$ members which cover all the societies. Call a family of societies "good" if each member appears at most $k$ times in the family. Take the maximal good family $\mathcal{F}$, and suppose it has $x$ societies in it. Clearly $x \le m$ by the problem condition. Now call a member "common" if it appears in exactly $k$ societies in $\mathcal{F}$, suppose there are $y$ common members. Now each society not in $\mathcal{F}$ must contain some common member of $\mathcal{F}$, otherwise we could add it to $\mathcal{F}$ which would contradict its maximality. Hence, if we pick the $y$ common members of $\mathcal{F}$, we cover all societies not in $\mathcal{F}$. Furthermore, each common member appears $k$ times in $\mathcal{F}$, so they cover altogether at least $y$ sets in $\mathcal{F}$. So if we take an arbitrary member of each of the remaining $x-y$ sets in $\mathcal{F}$, we will cover $\mathcal{F}$ as well. Hence we can use exactly $x\le m$ members to cover all the societies, as desired.