First, we will prove the following lemma:

Lemma. Given a positive integer $n$. Prove that the system of linear equations with $n$ variables $(x_1, x_2, \dots, x_n)$
$$\left\{ \begin{array}{l} a_{11}x_1 + \cdots + a_{1n}x_n = b_1, \\ \cdots \\ a_{n1}x_1 + \cdots + a_{nn}x_n = b_n \end{array} \right. \qquad (4.1)$$
has exactly one solution if the associated homogeneous system (which means $b_1 = \cdots = b_n = 0$) has only one solution $x_1 = \cdots = x_n = 0$.

Proof. Assume that both of $(x_1, \dots, x_n)$ and $(y_1, \dots, y_n)$ are the solutions of this system, we have
$$\left\{ \begin{aligned} & a_{11}(x_1 - y_1) + \cdots + a_{1n}(x_n - y_n) = 0, \\ & \cdots \\ & a_{n1}(x_1 - y_1) + \cdots + a_{nn}(x_n - y_n) = 0 \end{aligned} \right.$$
hence by the given condition, we obtain that $x_1 - y_1 = x_2 - y_2 = \dots = x_n - y_n = 0$, which means the system has at most one solution.

We will prove this system always has a solution by induction for $n$. It is obvious for $n = 1$. Assume that the lemma is proved for $n-1$. It is clear that if $a_{ij} = 0$ for all pairs $(i, j)$ then the associated homogeneous system has infinite solutions, hence there must exist $a_{ij} \neq 0$. Without loss of generality, assume that $a_{nn} \neq 0$. The system can be rewritten as follows
$$\left\{ \begin{aligned} & \sum_{i=1}^{n-1} \left( a_{1i} - a_{ni} \frac{a_{1n}}{a_{nn}} \right) x_i = b_1 - b_n \frac{a_{1n}}{a_{nn}}, \\ & \sum_{i=1}^{n-1} \left( a_{2i} - a_{ni} \frac{a_{2n}}{a_{nn}} \right) x_i = b_2 - b_n \frac{a_{2n}}{a_{nn}}, \\ & \cdots \\ & \sum_{i=1}^{n-1} \left( a_{n-1,i} - a_{ni} \frac{a_{n-1,n}}{a_{nn}} \right) x_i = b_{n-1} - b_n \frac{a_{n-1,n}}{a_{nn}}, \\ & \sum_{i=1}^{n} a_{ni} x_i = b_n \end{aligned} \right.$$
Clearly, if the system
$$\left\{ \begin{aligned} & \sum_{i=1}^{n-1} \left( a_{1i} - a_{ni} \frac{a_{1n}}{a_{nn}} \right) x_i = 0, \\ & \sum_{i=1}^{n-1} \left( a_{2i} - a_{ni} \frac{a_{2n}}{a_{nn}} \right) x_i = 0, \\ & \cdots \\ & \sum_{i=1}^{n-1} \left( a_{n-1,i} - a_{ni} \frac{a_{n-1,n}}{a_{nn}} \right) x_i = 0 \end{aligned} \right.$$
has a solution $(y_1, y_2, \dots, y_{n-1}) \neq (0, 0, \dots, 0)$ then the homogeneous system with $n$ variables $(x_1, x_2, \dots, x_n)$
$$\left\{ \begin{array}{l} \displaystyle \sum_{i=1}^{n-1} \left( a_{1i} - a_{ni} \frac{a_{1n}}{a_{nn}} \right) x_i = 0, \\ \displaystyle \sum_{i=1}^{n-1} \left( a_{2i} - a_{ni} \frac{a_{2n}}{a_{nn}} \right) x_i = 0, \\ \vdots \\ \displaystyle \sum_{i=1}^{n-1} \left( a_{n-1,i} - a_{ni} \frac{a_{n-1,n}}{a_{nn}} \right) x_i = 0, \\ \displaystyle \sum_{i=1}^{n} a_{ni} x_i = 0 \end{array} \right.$$
has a root $(y_1, y_2, \dots, y_{n-1}, 0)$, which is a contradiction. Thus, applying the assumption for $n-1$, the system
$$\left\{ \begin{array}{l} \displaystyle \sum_{i=1}^{n-1} \left( a_{1i} - a_{ni} \frac{a_{1n}}{a_{nn}} \right) x_i = b_1 - b_n \frac{a_{1n}}{a_{nn}}, \\ \displaystyle \sum_{i=1}^{n-1} \left( a_{2i} - a_{ni} \frac{a_{2n}}{a_{nn}} \right) x_i = b_2 - b_n \frac{a_{2n}}{a_{nn}}, \\ \vdots \\ \displaystyle \sum_{i=1}^{n-1} \left( a_{n-1,i} - a_{ni} \frac{a_{n-1,n}}{a_{nn}} \right) x_i = b_{n-1} - b_n \frac{a_{n-1,n}}{a_{nn}} \end{array} \right.$$
has exactly one solution $(z_1, \dots, z_{n-1})$ and note that
$$x_n = \frac{b_n - \sum_{i=1}^{n-1} a_{ni} b_i}{a_{nn}},$$
which implies that the lemma is also true for $n$. $\square$

Back to our problem, let $a_1, a_2, \dots, a_{2019}$ be the remaining numbers on 2019 faces and denote $a_{2020} = 4, a_{2021} = 26, a_{2022} = 2022$. Next, we write $b_{i,j} = 1$ if the face containing $a_i$ has a common edge with the face containing $a_j$, otherwise we write $b_{i,j} = 0$. Denote
$$b_{ii} = - \sum_{j=1, j \neq i}^{n} b_{ij}.$$
By the given conditions, we have the following system
$$\left\{ \begin{array}{l} \displaystyle \sum_{j=1}^{2019} b_{1,j}a_j = -4b_{1,2020} - 26b_{1,2021} - 2022b_{1,2022}, \\ \\ \displaystyle \sum_{j=1}^{2019} b_{2,j}a_j = -4b_{2,2020} - 26b_{2,2021} - 2022b_{2,2022}, \\ \\ \vdots \\ \\ \displaystyle \sum_{j=1}^{2019} b_{2019,j}a_j = -4b_{2019,2020} - 26b_{2019,2021} - 2022b_{2019,2022}. \end{array} \right.$$
Applying the lemma, it is clear that we only need to prove the system
$$\left\{ \begin{array}{l} \displaystyle \sum_{j=1}^{2019} b_{1,j}a_j = 0, \\ \\ \displaystyle \sum_{j=1}^{2019} b_{2,j}a_j = 0, \\ \\ \vdots \\ \\ \displaystyle \sum_{j=1}^{2019} b_{2019,j}a_j = 0, \end{array} \right.$$
has exactly one solution $a_1 = a_2 = \dots = a_{2019} = 0$. Assume that this system has another solution, which means there exists $j$ such that $a_j \neq 0$. Without loss of generality, assume that
$$a_1 = \max\{a_i : 1 \leq i \leq 2019\} > 0.$$
We observe that
$$M \left( \sum_{i=2}^{2019} b_{1,i} \right) \geq \sum_{i=2}^{2019} b_{1,i} a_i = a_1 \left( \sum_{i=2}^{2022} b_{1,i} \right) \geq M \left( \sum_{i=2}^{2019} b_{1,j} \right),$$
which means all the equalities must attain, or $a_i = M$ if $b_{1,i} = 1$ and
$$b_{1,2020} = b_{1,2021} = b_{1,2022} = 0.$$
Similarly, we obtain that for every $a_i = M$ then all the faces that have a common edge with $a_i$ contain $M$ and the face that contains $a_i$ has no common edge with the faces that contain $a_{2022}, a_{2021}$ and $a_{2020}$. Denote $A = \{i \ge 2019 : a_i = M\}, B = \{1, 2, \dots, 2022\} \setminus A$. Clearly, there exists $i \in A, j \in B$ such that the face containing $a_i$ has a common edge with the face containing $a_j$, which is a contradiction. Hence, the assumption is wrong, which means there exists $a_1$ such that $a_1 < 0$. Consider the solution
$$(c_1, c_2, \dots, c_{2019}) = (-a_1, -a_2, \dots, -a_{2019}),$$
we have a solution with the biggest number positive, which is a contradiction. Thus, $a_i = 0$ for all $i$ from 1 to 2019 is the only solution of that homogeneous system. Applying the lemma, the system has exactly one solution. $\square$