Let's remind Helly's theorem: if in a finite set of convex sets of points on a plane each three intersect, then all of them intersect.

Let's prove that $n = 9m$ satisfies the problem statement. Let $X$ be an arbitrary set of $9m$ different points on a plane, and $Y$ — the set of subsets of $X$ of size $6m + 1$.

Suppose that there exist such $A, B, C \in Y$ that their intersection is empty. Let's enumerate all points in $X$ by numbers from $1$ to $9m$. Let's write down on a sheet of paper the numbers of points in $A$, then the numbers of points in $B$, and after that the numbers of points in $C$. We wrote $|A| + |B| + |C| = 18m + 3$ numbers in total. Since these sets do not intersect, then we couldn't write any number more than twice. Thus, we wrote no more than $2 \cdot 9m = 18m$ numbers — a contradiction. Therefore, any three elements of $Y$ intersect.

Since the convex hull of a set of points contains the set itself, then the convex hulls of any three elements of $Y$ intersect. According to Helly's theorem, the convex hulls of all elements of $Y$ share some common point $O$.

Let's prove the following lemma: if the convex hull of a finite set of points $Z$ contains some point $P$, then there exists such $W \subseteq Z$ that $|W| \le 3$ and the convex hull of $W$ contains $P$. By definition of convex hull, there exists a convex polygon with the set of vertices $V \subseteq Z$ (possibly, degenerate) that contains $P$. If $|V| \le 3$, then $V$ works as $W$. Otherwise, let's perform an arbitrary triangulation of the polygon with vertices in $V$. Point $P$ has to lie in at least one of the obtained triangles. The set of vertices of such triangle works as $W$.

Suppose we have a bag into which we can put non-empty subsets of $X$. Let's denote the following operation, which modifies $X$ and $Y$: take any $A \in Y$. Since the convex hull of $A$ contains $O$, then, according to the lemma, there exists such $B \subseteq A$ that $|B| \le 3$ and the convex hull of $B$ contains $O$. Let's put $B$ into our bag (obviously, $B$ is non-empty), delete elements of $B$ from $X$, and delete sets from $Y$ that contain element of $B$.

After one such operation the size of $X$ decreases by at most three, and $Y$ remains non-empty as long as $|X| \ge 6m + 1$. Therefore, we can perform the operation at least $m$ times. Let's perform it exactly $m$ times. Distribute the remaining elements of $X$ randomly among the sets in the bag.

So, the sets in the bag constitute a partition of the initial set of points into $m$ non-empty sets and the convex hull of each of them contains point $O$, which is what we wanted.

Let's prove by induction on $m$ that any finite set of at least $4m^2$ different points on a plane can be partitioned into $m$ non-empty sets, convex hulls of which intersect. Obviously, the claim holds for $m=1$. Assume that it holds for $m=k-1$, where $k \ge 2$. Let's prove that it also holds for $m=k$. Let's consider an arbitrary finite set $X$ consisting of at least $4k^2$ different points on a plane. Let $Y$ be the subset of points of $X$ that lie on the border of the convex hull of $X$.

If $|Y| < 4k$, then $|X \setminus Y| > 4k^2 - 4k > 4(k-1)^2$. By the induction hypothesis, $X \setminus Y$ can be partitioned into $k-1$ non-empty sets, convex hulls of which intersect. If we add $Y$ to these $k-1$ sets, then we would get $k$ sets, convex hulls of which intersect (since the convex hull of $Y$ contains all points from $X \setminus Y$).

If $|Y| \ge 4k$, then there are two cases.

If all points from $Y$ lie on the same line, then all points from $X$ lie on the same line. Let's draw a coordinate axis along this line and denote the points from $X$ by $A_1, A_2, \dots, A_{|X|}$ in the order of increasing coordinates. Since $4k^2 > 2k$, then the following partition works:
$$X = \{A_1, A_{|X|}\} \cup \{A_2, A_{|X|-1}\} \cup \dots \cup \{A_{k-1}, A_{|X|-k+2}\} \cup \{A_k, A_{k+1}, \dots, A_{|X|-k+1}\}$$

Otherwise points of $Y$ lie on the border of some non-degenerate convex polygon. Denote the points from $Y$ in clockwise order by
$$A_1, A_2, \dots, A_k, B_k, B_{k-1}, \dots, B_1, C_1, C_2, \dots, C_k, D_k, D_{k-1}, \dots, D_1, E_1, E_2, \dots, E_{|Y|-4k}.$$
Let
$$Z = \{A_k, B_k, C_k, D_k, E_1, \dots, E_{|Y|-4k}\} \cup (X \setminus Y).$$
Let's prove that the following partition works:
$$X = \left( \bigcup_{i=1}^{k-1} \{A_i, B_i, C_i, D_i\} \right) \cup Z.$$
It is enough to prove the key assertion: convex hulls of
$$\{A_1, B_1, C_1, D_1\}, \{A_2, B_2, C_2, D_2\}, \dots, \{A_k, B_k, C_k, D_k\}$$
intersect. Denote the convex hull of a set of points $M$ by $f(M)$. Let
$$T \in A_1B_1 \cap A_kD_k,$$
$$H_i = f(\{A_1, A_2, \dots, A_i, B_i, B_{i-1}, \dots, B_1, C_1, C_2, \dots, C_i, D_i, D_{i-1}, \dots, D_1\})$$
and
$$V_i = f(\{A_i, A_{i+1}, \dots, A_k, B_k, B_{k-1}, \dots, B_i, C_i, C_{i+1}, \dots, C_k, D_k, D_{k-1}, \dots, D_i\}).$$
Then for each $1 \le i \le k$ holds
$$T \in A_1 B_1 \subseteq H_i \text{ and } T \in A_k D_k \subseteq V_i.$$
Therefore,
$$T \in \bigcap_{i=1}^{k} (H_i \cap V_i) = \bigcap_{i=1}^{k} f(\{A_i, B_i, C_i, D_i\}),$$
Q.E.D.