Solution:

Let us begin with some preliminaries. In the solution to follow, we treat points freely as vectors, e.g. writing $n A$ to mean the point whose coordinates are $n$ times the coordinates of $A$, or $A+B$ to mean the point which is the coordinate-wise sum of $A$ and $B$.
A basic result from vector geometry, which we will assume, states that given any three noncollinear points $A, B, C$ in the plane, every point $Q$ may be represented in the form $r A+s B+t C$ for unique $r, s, t$ satisfying $r+s+t=1$. Moreover, $Q$ is in the interior of $\triangle A B C$ if and only if such $r, s, t$ are all positive. In similar fashion, any point on the line through $A$ and $B$ can be expressed as $r A+s B$ with $r+s=1$, and lies between $A$ and $B$ if and only if $r, s>0$.
We will also make repeated use of Pick's Theorem, which we now state without proof. This theorem asserts that a lattice polygon (a polygon whose vertices are lattice points) has area equal to
$$i+\frac{1}{2} b-1$$
where $i$ and $b$ are the number of lattice points on the polygon's interior and boundary, respectively. Thus (for instance), the triangle $T$ described in the problem must have area $4+\frac{1}{2}(3)-1=\frac{9}{2}$.
Now we are ready to begin the solution. With no loss of generality, let us assume $T$ has one vertex at the origin $O$, which we identify with the zero vector. Call the other two vertices $A$ and $B$.
Of the four lattice points in the interior of $T$, let $P$ be the point closest to line $O A$. It follows that there are no lattice points lying inside $\triangle O P A$ or on its boundary, other than $O, P, A$ themselves, since any such point would be closer than $P$ to line $O A$. Therefore, by Pick's Theorem, $\triangle O P A$ has area $\frac{1}{2}$.

Lemma: Every lattice point $Q$ can be expressed in the form $n P+k A$ for some pair of integers $(n, k)$. Moreover, when $Q$ is expressed in such form, we have $n=2[O Q A]$. (The brackets represent area.)

Proof. Let $Q$ be a lattice point. By Pick's Theorem, $[O Q A]=\frac{n}{2}$ for some integer $n$. Thus $[O Q A]=n \cdot[O P A]$. By the base-height formula for triangle area, it follows that $Q$ is on the line parallel to line $O A$ that passes through the point $n P$. Thus $Q=n P+k A$ for some real $k$, where $k A$ is a lattice point.
We assert that $k$ is an integer. Indeed, if $\{k\}$ denotes the fractional part of $k$, then $\{k\} A=k A-\lfloor k\rfloor A$ is a lattice point which lies on segment $O A$, part of the boundary of $T$. Since $T$ has no lattice points on its boundary other than its vertices, we must have $\{k\}=0$. This completes the proof of the lemma.

Let us return to the main problem. As already noted, $[T]=[O B A]=\frac{9}{2}$. Thus by the lemma, $B=9 P- k A$ for some integer $k$ (the minus sign in the expression is not a typo, but a deliberate convenience for what follows). Rearranging, and using the fact that $O$ is the zero vector, we have $P=\frac{k}{9} A+\frac{1}{9} B+\frac{8-k}{9} O$. Since $P$ is in the interior of $T$, we have $0<k<8$. We will consider the possible values of $k$ in turn.

If $k \equiv 0(\bmod 3)$, then $\frac{1}{3} B=3 P-\frac{k}{3} A$ is a lattice point lying on segment $O B$. This contradicts the specification of $T$ as having no lattice points on its sides.
If $k \equiv 2(\bmod 3)$, then $\frac{1}{3} B+\frac{2}{3} A=3 P-\frac{k-2}{3} A$ is a lattice point lying on $A B$, similarly yielding a contradiction.
The remaining possibilities are $k=1,4,7$.

If $k=1$, then the interior of $T$ contains in its interior the four collinear lattice points
$$P=\frac{1}{9} A+\frac{1}{9} B+\frac{7}{9} O, \quad 2 P=\frac{2}{9} A+\frac{2}{9} B+\frac{5}{9} O, \quad 3 P=\frac{3}{9} A+\frac{3}{9} B+\frac{3}{9} O, \quad 4 P=\frac{4}{9} A+\frac{4}{9} B+\frac{1}{9} O$$
If $k=4$, then the interior of $T$ contains in its interior the four collinear lattice points
$$P=\frac{4}{9} A+\frac{1}{9} B+\frac{4}{9} O, \quad 3 P-A=\frac{3}{9} A+\frac{3}{9} B+\frac{3}{9} O, \quad 5 P-2 A=\frac{2}{9} A+\frac{5}{9} B+\frac{2}{9} O, \quad 7 P-3 A=\frac{1}{9} A+\frac{7}{9} B+\frac{1}{9} O .$$
If $k=7$, then the interior of $T$ contains in its interior the four collinear lattice points
$$P=\frac{7}{9} A+\frac{1}{9} B+\frac{1}{9} O, \quad 2 P-A=\frac{5}{9} A+\frac{2}{9} B+\frac{2}{9} O, \quad 3 P-2 A=\frac{3}{9} A+\frac{3}{9} B+\frac{3}{9} O, \quad 4 P-3 A=\frac{1}{9} A+\frac{4}{9} B+\frac{4}{9} O$$
Thus, the four lattice points inside $T$ are collinear in every case, as desired.

Solution:

We assume the same basic facts about vectors as in the previous solution, as well as Pick's Theorem and the determinant formula for the area of a parallelogram.
Let $T$ have vertices $A=(x_{1}, y_{1}), B=(x_{2}, y_{2})$, and $C=(x_{3}, y_{3})$. We know that
$$\left|\begin{array}{ll} x_{2}-x_{1} & x_{3}-x_{1} \\ y_{2}-y_{1} & y_{3}-y_{1} \end{array}\right|=2[A B C]=9$$
Consider equation (1) modulo $3$, that is, over the field $\mathbb{Z} / 3 \mathbb{Z}$. In this setting, the determinant is zero, so the vectors $\mathbf{u}=(x_{2}-x_{1}, y_{2}-y_{1})$ and $\mathbf{v}=(x_{3}-x_{1}, y_{3}-y_{1})$ are linearly dependent. If either of these vectors is zero $(\bmod 3)$, that is, or if they are equal, then the trisection points of a side of $T$ are lattice points, which contradicts the problem statement. Thus $\mathbf{u}, \mathbf{v} \neq \mathbf{0}$ and $\mathbf{u}=-\mathbf{v}$.
An immediate consequence is that $(x_{1}+x_{2}+x_{3}, y_{1}+y_{2}+y_{3})=\mathbf{u}+\mathbf{v}+3(x_{3}, y_{3})=\mathbf{0}$ over $\mathbb{Z} / 3 \mathbb{Z}$, with the result that the centroid, $G=\frac{1}{3}(A+B+C)$, is a lattice point.
Now consider $\triangle A B G$, whose area is $\frac{1}{3}[A B C]=\frac{3}{2}$. By Pick's Theorem, $\triangle A B G$ has either
- one lattice point in its interior and none on its boundary (besides vertices), or
- two lattice points on its boundary.

Case 1: $\triangle A B G$ has a lattice point in its interior and none on its boundary. In this case, a repetition of the preceding $(\bmod 3)$ argument shows that the centroid $G_{1}$ of $\triangle A B G$ is a lattice point. In this case, $G_{1}+k(G-G_{1})$ for $k=0,1,2,3$ are four collinear lattice points inside $T$.

Case 2: $\triangle A B G$ has two lattice points on its boundary. Note that if at least two lattice points occur on a line, then the lattice points on that line occur at regular intervals. Thus the two lattice points on the boundary of $\triangle A B G$ are either the midpoints of $A G$ and $B G$ or the trisection points of one of these sides (say, $A G$). In the two cases, if we extend side $A G$ beyond $G$, the next lattice point occurring on the extension is respectively either on $T$ (at the midpoint of side $B C$), which is a contradiction, or inside $T$, being then the fourth collinear lattice point inside $T$. So we are finished.