We claim the answer is 1000. This is achieved by the above construction. Let $a_{ij}$ be the number of "connections" between persons $i$ and $j$, $i<j$ (if person i invites person j but person j does not invite person i, that is 1 connection, and if they invite each other, then it is 2 connections). We have $a_{ij}=\{0,1,2\}$ and $i$ and $j$ are friends if and only if $a_{ij}=2$. Note that $\sum a_{ij}=1000\times 2000$ because each of 2000 people are "connected with" 1000 other people. Also, note that there are only $\binom{2000}{2}=1000\times 1999$ possible pairs of $i$ and $j$, which is 1000 less than $1000\times 2000$. Now, assume for contradiction that the number of $i, j$ such that $a_{ij}=2$ is less than 1000. We have $\sum a_{ij}\le 999(2)+(1000\times 1999-999)(1)+0(0)=1000\times 2000-1<1000\times 2000$, a contradiction. Therefore, there must be at least 1000 pairs $i, j$ such that $a_{ij}=2$ and we are done.