We claim that the least possible by number of inhabitants who will become sad is $674$. To achieve this, the first $2019$ inhabitants form groups of $3$, and everyone chooses the other two members of their group to send a Christmas card. Then, in each group of $3$ , atleast $2$ receive a Christmas Card from someone else, which implies at most one person from each group will be left sad. Including the other person who is not included in any group this gives exactly, \[\frac{2019}{3}+1=673+1=674\]sad inhabitants. Now, we show that the company can always force atleast $674$ inhabitants to be sad. We do this by showing the more general result that in a town consisting of $n$ inhabitants, the company can ensure atleast $\lceil \frac{n}{3} \rceil$ sad inhabitants. We say a pair of inhabitants $A$ and $B$ are cofriends if some inhabitant $C$ picks $A$ and $B$ as his two recipients of Christmas cards. Note that it suffices to show that in a town consisting of $n$ inhabitants there are atleast $\lceil \frac{n}{3} \rceil$ inhabitants who are pairwise not cofriends of each other, since then the mailing company can mail the Christmas card to the cofriend of each of these inhabitants (which are guaranteed to be distinct) so we have atleast $\lceil \frac{n}{3} \rceil$ sad inhabitants. We prove this claim via induction. Claim : In a town consisting of $n$ inhabitants, and at most $n$ pairs of cofriendships, there exists a set $\mathcal{S}$ of atleast $\lceil \frac{n}{3} \rceil$ inhabitants such that no pair of inhabitants in $\mathcal{S}$ are cofriends of each other. Proof : The (degenerate) cases where $n=1,2$ and $3$ are immediate. Now, assume the claim is true for some $k\ge 3$, we attempt to show that it also holds for $n=k+3$. Note that there exactly $n$ pairs of cofriendships. Now, there are $2n$ instances of some inhabitant being in a cofriendship. We split into two cases. Case 1 : Each inhabitant is in exactly $2$ cofriendships. Consider some inhabitant $X$ and his two cofriends $Y$ and $Z$. Consider the $n-3=k$ person set including all inhabitants besides $X,Y$ and $Z$. There by the inductive hypothesis exists a set of at least $\lceil \frac{k}{3} \rceil$ inhabitants who are not cofriends of each other. Adding only $X$ to this set then gives us a set of $\lceil \frac{n}{3}\rceil$ set of inhabitants of which no two are cofriends of each other, completing the inductive step. Case 2 : There exists some inhabitant $P$ in 1 or less cofriendships (and in turn some inhabitant $Q$ in 3 or more cofriendships). Consider the set of $k$ people excluding $P$ , his (possible) cofriend $P'$ (if $P$ doesn't have a cofriend consider some random person besides $P$ and $Q$ to be $P'$), and $Q$. Now, in the remaining set of $k$ people, and at most $k$ cofriendships (removing $Q$ deletes 3 or more pairs of cofriends) we have a set of at least $\lceil \frac{k}{3} \rceil$ people who are not cofriends of each other. Then, add only $P$ to this set, so we have a set of atleast $\lceil \frac{n}{3} \rceil$ inhabitants who are pairwise non-cofriends of each other. This completes the induction, and finishes the proof.