Bump again. I'll put graph formulation here: Given is a graph $G$, such that every two vertices have even number of common neighbors. We want to orient some of the edges, so that each "odd vertex" has outdeg=1, and each "even vertex" has outdeg=0 or 1, and each vertex has indeg=0 or 1. Prove that the number of ways to do so is odd.

Well, this is embarrassing. Solution by me and "Wilbert" on the OTIS server independently. With help from AwesomeYRY. Ignoring the "every resident received no more than one postcard" condition, there are an odd number of total assignments (each person has an odd number of choices). We claim the set $S$ of assignments where at least one pair of residents sends cards to the same person has even size. We provide an algorithm by which we reduce $S$ to the empty set. 1. If $S$ is empty we are done. 2. If $S$ is not empty then take an assignment in $S$ and some pair $A,B$ sending cards to the same person $C$ in the assignment. Remove another assignment that is identical to this assignment except for the fact that $A,B$ sends cards to a person $D \ne C.$ We repeat these two steps as many times as necessary until $S$ is empty. Since each pair has an even number of common friends, we'll always be able to subtract one pair at a time. Thus $S$ has even size. $\square$ @below I made a mistake when typing up my solution, I believe this now works. It is similar to a1267ab's approach.

I'm not sure what do you mean by squareman wrote: Remove all assignments where this pair of residents send cards to the same person. I suppose that you find a pair of residents $x,y$, and remove all the assignments currently in $S$ such that $x,y$ send cards to the same person. But it may happen that some assignments are removed before, say, $x,y,z$ both send cards to $w$, and we have removed all assignments with $y,z$ sending cards to the same person in previous round, so it no longer appears in $S$. Thus for each common friend $w$ of $x,y$, after making $x,y$ send cards to $w$, the restrictions on other people may not be the same (say, in the above example, $z$ cannot send card to $w$, but for some other $w'$ he can), so the number of assignments remained in $S$ such that $x,y$ send cards to $w$ varies, depending on $w$. I'm not sure we can conclude the number of assignments removed is even in this way. Maybe my understanding is incorrect or missing something. @above I believe that this procedure squareman wrote: If $S$ is not empty then take an assignment in $S$ and some pair $A,B$ sending cards to the same person $C$ in the assignment. Remove another assignment that is identical to this assignment except for the fact that $A,B$ sends cards to a person $D \ne C.$ has the same problem as I pointed above, because some assignments are removed before (say, $E$ sends card to $D$ in the assignment). a1267ab's works fine, but I don't think it has any similarities with your solution. Maybe I am too dumb to understand your solution, which must be exquisite as it obtained 12 upvotes! (Sadly, these 12 people who understand the solution don't want to clarify for me )

I think @above is right. You have to consider for how many pairs $x, y$ each assignment is counted.

We can solve this problem in a similar way to USAMO 2018/6. Consider the set $S$ of bad assignments as squareman did. Let two assignments in $S$ be connected if one assignment can be obtained from the other by the following procedure: Take some disjoint pairs of people who sent postcards to the same person. Change each chosen pair to send postcards to a different common friend. Each assignment of $S$ is connected to an odd number of assignments (this is the same as the USAMO problem so I won't show the details), so by the handshaking lemma, $S$ has even size. squareman wrote: We provide an algorithm by which we reduce $S$ to the empty set. 1. If $S$ is empty we are done. 2. If $S$ is not empty then take an assignment in $S$ and some pair $A,B$ sending cards to the same person $C$ in the assignment. Remove another assignment that is identical to this assignment except for the fact that $A,B$ sends cards to a person $D \ne C.$ We repeat these two steps as many times as necessary until $S$ is empty. Since each pair has an even number of common friends, we'll always be able to subtract one pair at a time. Thus $S$ has even size. $\square$ This is wrong, and very similar to a common fakesolve of the USAMO problem.