Solution:

We claim that all rooms must be on the same hall. Suppose for contradiction there is a room other than room $1$ that is the first room on its hall. We will show by induction that its room number must be congruent to $2^{2k-1} \pmod{3 \cdot 2^{2k-1}}$ for every positive integer $k$, and therefore must be infinitely big.

First, note that room $2n$ is on the same hall as room $3n$ for every $n$, since both are connected to room
$$3 \cdot 2n + 1 = 2 \cdot 3n + 1 = 6n + 1.$$
Thus, no room that is a multiple of $3$ is first on its hall. Also, no room that is odd or that is $1 \pmod{3}$ besides room $1$ is first on its hall, since if $n$ is odd room $n$ is on the same hall as room $(n-1)/2$, and if $n$ is $1 \pmod{3}$ room $n$ is on the same hall as room $(n-1)/3$.

Now, for the base case, we claim that a room other than room $1$ that is first on its hall must be $2 \pmod{6}$. To see this, note that the room number cannot be $1, 3$, or $5 \pmod{6}$, since then it would be odd; it cannot be $0 \pmod{6}$, since then it would be a multiple of $3$; and it cannot be $4 \pmod{6}$, since then it would be $1 \pmod{3}$. Thus, it must be $2 \pmod{6}$.

For the inductive step, suppose our room number can be written as $3 \cdot 2^{2k-1} \cdot m + 2^{2k-1}$ for nonnegative integers $m$ and $k$. We claim first that $m$ is odd. Thus, suppose $m$ is even.

By repeatedly applying the rule that room $2n$ is on the same hall as room $3n$, our room is on the same hall as room
$$\left(\frac{3}{2}\right)^{2k-1}\left(3 \cdot 2^{2k-1} \cdot m + 2^{2k-1}\right) = 3^{2k} \cdot m + 3^{2k-1}.$$
Since we are assuming $m$ is even, this room number is odd, so it is on the same hall as room
$$\frac{3^{2k} \cdot m + 3^{2k-1} - 1}{2}.$$
This room number is $1 \pmod{3}$, so it is on the same hall as room
$$\frac{\frac{3^{2k} \cdot m + 3^{2k-1} - 1}{2} - 1}{3} = \frac{3^{2k-1} \cdot m + 3^{2k-2} - 1}{2}.$$
Repeatedly applying the rule that room $n$ is on the same hall as room $(n-1)/3$, we find that this room is on the same hall as room
$$\frac{3^{i} \cdot m + 2^{i-1} - 1}{2}$$
for every positive integer $i$. In particular, for $i=1$, it is on the same hall as room $3m/2$. This is a smaller room number than the one we started with, contradicting the assumption that our room was the first on its hall. We conclude that $m$ must be odd.

Thus, we can write $m = 2j + 1$ for some $j$, so our room number is now
$$3 \cdot 2^{2k-1}(2j+1) + 2^{2k-1} = 3 \cdot 2^{2k} \cdot j + 2^{2k+1}.$$
We now claim that $j$ is even, which will complete our induction.

Thus, suppose $j$ is odd. As before, we repeatedly apply the rule that rooms $2n$ and $3n$ are on the same hall to find that this room is on the same hall as room
$$\left(\frac{3}{2}\right)^{2k}\left(3 \cdot 2^{2k} \cdot j + 2^{2k+1}\right) = 3^{2k+1} \cdot j + 2 \cdot 3^{3k}.$$
Since we are assuming $j$ is odd, this number is odd, so it is on the same hall as room
$$\frac{3^{2k+1} \cdot j + 2 \cdot 3^{3k} - 1}{2}.$$
This room number is $1 \pmod{3}$, and similarly to before, we find by repeatedly applying the rule that rooms $n$ and $(n-1)/3$ are connected that our room is on the same hall as room
$$\frac{3^{i} \cdot j + 2 \cdot 3^{i-1} - 1}{2}$$
for each positive integer $i$. In particular, it is on the same hall as room $(3j+1)/2$, which is smaller than our starting room number. This contradicts the assumption that $j$ was odd, so $j$ must be even. This completes our proof.