Actually this one - as I also learned after the contest - is a special case of a known theorem which is called Friendship Theorem. Take a look at it from here

I got that the only possible graph is that one vertex has $ 2008$ edges,and other $ 2008$ vertices form $ 1004$ egdes. So the answer is $ 2008-2=2006$ Is it right?

Yes, it's correct.

My solution: Take the person with the maximal number of friends. Call him $ A$. Assume all of $ A$'s friends are $ A_{1},A_{2},\cdots,A_{m}$, respectively. For $ 1\le i\le m$, since $ A$ and $ A_{i}$ have exactly one friend, which must be one of $ A_{1},A_{2},\cdots,A_{m}$. Therefore $ A_{i}$ has exactly one friend among $ A_{1},A_{2},\cdots,A_{m}$. So $ m$ is even. Assume $ m=2k$. W.l.o.g. we can assume that for $ 1\le i\le k$, $ A_{2i-1}$ and $ A_{2i}$ are friends. Since $ A$ is chosen arbitrarily, we have that everyone has an even number of friends. Now it suffices to prove that $ m=2008$, which is equivalent to that there are no more people other than $ A,A_{1},A_{2},\cdots,A_{m}$. Assume the opposite, for any person $ B$ other than $ A,A_{1},A_{2},\cdots,A_{m}$. Since $ A$ and $ B$ have exactly one friend, $ B$ has exactly one friend among $ A_{1},A_{2},\cdots,A_{m}$. So we can divide other persons into $ m=2k$ sets according to which of $ A_{1},A_{2},\cdots,A_{m}$ is his friend. More specific, for $ 1\le i\le m$, denote set $ F_{i}=\{ x|x\notin \{A,A_{1},A_{2},\cdots,A_{m}\} \text{ and x is a friend of }A_{i} \}$ and $ S_{i}=|F_{i}|$. It is trivial that $ k\ge 2$, so $ A_{3}$ exists. Due to the opposite assume $ F_{i}$ are not all empty, w.l.o.g. we assume $ F_{1}$ is not empty and take an element $ B_{1}$ from it. For $ 3\le i\le m$, $ B_{1}$ and $ A_{i}$ have exactly one friend, which belongs to $ F_{i}$. Therefore $ B_{1}$ has exactly one friend in $ F_{i}$ for $ 3\le i\le m$. So there exists a bijection between $ F_{1}$ and $ F_{i}$ for $ 3\le i\le m$, so $ S_{1}=S_{3}=\cdots =S_{m}$. Due to the symmetry of all $ S_{i}$ we have all $ S_{i}$ are the same number. Assume it is $ x$. Because all of $ A_{1}$'s friends are $ A$, $ A_{2}$ and all $ x$ elements in $ F_{1}$, so $ A_{1}$ has exactly $ x+2$ friends and $ x$ is an even number. Therefore there are $ 1+m+mx=m(x+1)+1$ people in total. Also the condition told us that there are $ 2009$ people in total. So $ m(x+1)+1=2009$, $ m(x+1)=2008$, $ k(x+1)=1004$. Now it is easy to get a contradiction. $ 1004=2^2*251$, and $ 251$ is a prime number. But $ x+1$ is an odd divisor of $ 1004$ that means it could only be either $ 1$ or $ 251$. If $ x+1=251$, then $ x=250$ and $ k=4$, which makes $ A$ has less friends than $ A_{1}$, a contradiction. If $ x+1=1$, then $ x=0$ also contradicts the hypothesis. Therefore the hypothesis is wrong and the only possible graph is which I describe in the former post. Q.E.D