Prove $\exists T\neq \emptyset,\sum\limits_{v\in V}C=0$

I have a set of points in the complex plane that the parameters of the real and imaginary values are integers $P=\{a+bi,\enspace a,b\in\mathbb{Z}\}$ so the set P makes up a grid. There is a second grid $V$ offsetted by $\frac{1+i}{2}$ and it is defined by $V=\{p+\frac{1+i}{2},\enspace p\in P\}$. An element of $P$ and $V$ are called $p$ and $v$ respectively, formally: $p\in P,\enspace v\in V$.

The point $v$ represents a cylinder which can be rotated clockwise or counterclockwise by $90^{\circ}$, we can map the rotations $\{\text{right},\text{up},\text{left},\text{down}\}$ to $\{0,1,2,3\}$. Every cylinder starts at rotation $0$.

A person starts in position $0$ or $0+0i$ and each step he moves up, down, left or right $1$ unit at a time. Whenever the person passes through two $v_1$ and $v_2$ next to each other, $v_1,v_2\in V$, the one on the left is rotated $90^{\circ}$ counterclockwise and the one on the right is rotated $90^{\circ}$ clockwise.

Question:

Is it possible to make a path so that the initial state is the same as the end state? If yes, what conditions should $T$ have?

My attempt:

There are infinitely many trivial solutions such as $T=(1,-1)$ or $T=(1,1,-1,-1)$ or $T=(i,i,i,-i,-i,-i)$ which doesn't count. The general form of these solutions is $T=(t_1,\dots, t_n,-t_1,\dots ,-t_n)$ or $T=(t_1,\dots, t_n,-t_n,\dots ,-t_1)$

The path $T=(t_1,\dots,t_n),\enspace t_i\in\{1,i,-1,-i\}$.

The position of the person after $n$ steps is given by $p_n=\sum\limits_{i=1}^{n}t_i$.

At any step $n$ the position of the cylinder in the left is given by $$V_L(n)=p_n+\frac{1+i}{2}t_{n+1}=\frac{1+i}{2}t_{n+1}+\sum_{i=1}^{n}t_i$$ and the one on the right by $$V_R(n)=p_n+\frac{1-i}{2}t_{n+1}=\frac{1-i}{2}t_{n+1}+\sum_{i=1}^{n}t_i$$

$$V_L:\mathbb{N_0}\rightarrow V ,\quad V_R:\mathbb{N_0}\rightarrow V$$ $C_L(v)$ and $C_R(v)$ tells us the number of times the path has passed from the left and right, respectively, in any given cylinder at position $v$. Formally, $$C_L:V\rightarrow\mathbb{N_0} ,\quad C_R:V\rightarrow\mathbb{N_0}$$ $$C_L(v)=|\{k\mid v=V_L(k),\enspace k=1,\dots,\vert T\vert-1\}|$$ $$C_R(v)=|\{k\mid v=V_R(k),\enspace k=1,\dots,\vert T\vert-1\}|$$

The total rotation of the cylinder, is given by the number of path segments in the left minus the ones on the right modulo $4$.

$$C(v)= (C_L(v)-C_R(v))\mod{4}$$ $$\text{or}\quad C= (C_L-C_R)\mod{4}$$

Now I have to prove

$$\Large\quad\exists T\neq \emptyset,\sum\limits_{v\in V}C=0$$

By experiment, any path $T=(t_1,\dots, t_n,t_1,\dots, t_n,t_1,\dots, t_n,t_1,\dots, t_n)$ is a valid solution for this problem, but I don't know how to get there or even if these are the only valid solutions. For this $T$,
$$\sum\limits_i t_i=0$$
Does this mean that for the solutions to be valid the only way is to make a path that starts and ends at the same point and go through it $4$ times one after another?
$$4C\equiv 4(C_L-C_R)\mod{4}$$ $$4C\equiv 0\mod{4}$$ I think this is why passing through a closed path 4 times resets everything, but is it the only solution? Is it possible for a path which doesn't end in the start position?