The answer is $0, 2019, 2018 \times 2019, 2019^2$.

Consider the graph $G(V, E)$, where the vertex set and edge set correspond respectively to the tent poles and the ropes. For sets $S, T \subset V$, let
$$E(S, T) = \left| \left\{ (s,t) \in E : s \in S, t \in T \right\} \right|,$$
that is, the number of edges connecting the two groups. To simplify notation, when $S$ consists of only a single point $a$, we abbreviate this set as $a$, and so on.

1. Lemma: For all distinct $a, b, c, d \in V$, we have
$$E(a, b) + E(c, d) = E(a, d) + E(b, c) = E(a, c) + E(b, d)$$
Proof: First divide $V \setminus \{a, b, c, d\}$ equally into two sets $S, T$. Taking the two banks of the river to be $(a, b, S)$ and $(c, d, T)$ respectively, then by the problem's hypothesis we have
$$E(a, c) + E(a, d) + E(b, c) + E(b, d) + E(a, T) + E(b, T) + E(c, S) + E(d, S) + E(S, T) = k \quad (1)$$
Taking the two banks of the river to be $(a, d, S)$ and $(c, b, T)$ respectively, then by the problem's hypothesis we have
$$E(a, c) + E(a, b) + E(d, c) + E(d, b) + E(a, T) + E(b, S) + E(c, S) + E(d, T) + E(S, T) = k \quad (2)$$
Taking the two banks of the river to be $(c, d, S)$ and $(a, b, T)$ respectively, then by the problem's hypothesis we have
$$E(c, a) + E(c, b) + E(d, a) + E(d, b) + E(a, S) + E(b, S) + E(c, T) + E(d, T) + E(S, T) = k \quad (3)$$

(3) – (1) gives
$$(E(a, S) - E(a, T)) + (E(b, S) - E(b, T)) = (E(c, S) - E(c, T)) + (E(d, S) - E(d, T))$$
Similarly we have
$$(E(a, S) - E(a, T)) + (E(d, S) - E(d, T)) = (E(c, S) - E(c, T)) + (E(b, S) - E(b, T))$$
Subtracting the two equations gives $E(b, S) - E(b, T) = E(d, S) - E(d, T)$.

$$E(a, b) + E(d, c) + E(b, S) + E(d, T) = E(a, d) + E(b, c) + E(b, T) + E(d, S)$$
Combined with the equation above, we get $E(a, b) + E(d, c) = E(a, d) + E(b, c)$, thus the first part of the Lemma is proved. By the symmetry of $a, b, c, d$, the second equality also holds.

- If for all $a, b, c \in V$, $E(a, b) = E(a, c) = 0$, then the graph has no edges at all, corresponding to $k = 0$.
- If for all $a, b, c \in V$, $E(a, b) = E(a, c) = 1$, then the original graph is a complete graph, corresponding to $k = 2019^2$.
- Suppose there exist $a, b, c \in V$ such that $E(a, b) = 1, E(a, c) = 0$. Then for all $d$, we have $E(a, b) + E(c, d) = E(a, c) + E(b, d)$, so $E(b, d) = 1, E(c, d) = 0$.
If $E(b, c) = 1$, then $b$ has an edge to every point. For all points $u, v, w$ other than $b$, we have $E(b, u) + E(v, w) = E(b, v) + E(u, w)$, hence $E(u, w)$ and $E(v, w)$ are either both 0 or both 1. Also, since $E(a, c) = 0$, $V \setminus \{b\}$ forms an empty graph, that is, $G$ is a complete bipartite graph $(1, 4037)$, corresponding to $k = 2019$.

If $E(b, c) = 0$, then $c$ has no edge to any point. By a similar discussion we can see that $V \setminus \{c\}$ forms a complete graph, that is, $G$ is the union of a 4037-complete graph and an isolated point, corresponding to $k = 2018 \times 2019$.