The idea of the weights used in the ADN definition of the principal symbol is quite natural in the context of [graded vector spaces](https://en.wikipedia.org/wiki/Graded_vector_space) or more generally [graded modules](https://en.wikipedia.org/wiki/Graded_ring#Graded_module).

In the equation $u = Dv$, $u=(u_1,\ldots,u_n)$ and $v=(v_1,\ldots,v_n)$ are functions valued in vector spaces, say $U$ and $V$ respectively. A decomposition of $U = \bigoplus_a U_a$ and $V = \bigoplus_b V_b$ where $a$ and $b$ range over the integers $\mathbb{Z}$ endows these spaces with an integer grading. For simplicity, let us suppose that the components $(u_i)$ and $(v_j)$ are with respect to a basis that is compatible with the grading, which is the same as saying that each of the vector components has a well-defined integer degree. Since our vectors have only finitely many components, only finitely many of the $U_a$ or $V_b$ vector spaces will be non-trivial (have positive dimension). The punchline will be that the degree $a_i$ of the component $u_i$ could be identified with the ADN weight $s_i = a_i$, while the degree $b_j$ of the component $v_j$ could be identified with the ADN weight $-t_j = b_j$.

There is a notion of a homogeneous map between graded vector spaces. Namely $T \colon V \to U$ is _homogeneous of degree $c$_ if $T(V_b) \subseteq U_{a=b+c}$. Written as a matrix $T = (T_{ij})$, only those components $T_{ij}$ can be non-zero for which $a_i - b_j = c$. Of course, any linear map can be written as a sum of homogeneous pieces, $T = \sum_c T_{(c)}$. When the components of $T$ are not just numbers but polynomials in $\xi_l$, conventionally the variables $\xi_l$ are all given weight 1 and the notion of homogeneity changes slightly. Namely, $T$ is _(polynomially) homogeneous of degree $c$_ if its components $T_{ij}$ are homogeneous polynomials, say of degrees $d_{ij}$, and are non-vanishing only when $a_i - b_j - d_{ij} = c$. An arbitrary linear operator with polynomial entries with can again be expanded in homogeneous pieces, $T = \sum_{c} T_{[c]}(\xi)$, where each polynomially homogeneous pieces can again be expanded as $T_{[c]}(\xi) = \sum_{|\alpha|=c} T_{(c-|\alpha|)} \xi^\alpha$ with respect to the first notion of homogeneity. What I ad-hoc called polynomial homogeneity coincides with the notion of homogeneity when we consider $U$ and $V$ are graded modules over the polynomial ring $\mathbb{R}[\xi]$, where the $\xi_l$ variables all have weight 1.

Now, suppose that some integer grading is fixed on $U$ and $V$ and you have a linear operator between them with polynomial coefficients that has a homogeneous expansion $T = \sum_{c\le d} T_{[c]}(\xi)$, then $T_{[d]}(\xi)$ is its _leading homogeneous term_ with respect to the grading. If you are free to adjust the grading then without loss of generality we can set $d=0$ by the shift $a_i \mapsto a_i + d$ or $b_i \mapsto b_i - d$. Now, getting back to a linear differential equation $u = Dv$, the components of the operator $D$ are polynomials in the variables $\xi_l = \partial/\partial x^l$. It will have a leading $\xi$-homogeneous term $D_{[d]}(\xi)$ of some degree $d$.

The conclusion is that the ADN weighted principal symbol $\sigma_{\text{pr},w}(\xi) = D_{[d]}$ is just the leading homogeneous term with respect to a grading given by the weights $a_i = s_i$, $b_j = -t_j$ and $d=0$. But as we have seen earlier, the $d=0$ condition is easily relaxed just by shifting the weights. What is interesting about the definition of the weighted principal symbol as the leading homogeneous term is that, for a fixed choice of weights, it is easy to see that it is well-defined and transforms covariantly in $\xi$ under changes of the $(x^l)$ coordinates (subleading terms in the transformation of highest order derivatives in each component of $D_{ij}$ only contribute to subleading terms $D_{[c]}$, while the highest derivatives transform covariantly).

The usual definition of the principal symbol can be recovered by setting all the weights to zero, $a_i = b_j = 0$. However, the simple example
$$ \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}
  = \begin{pmatrix} \Delta^3 & 0 & 0 \\ 0 & \Delta^2 & 0 \\ 0 & 0 & \Delta \end{pmatrix}
    \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}
$$
shows that such a default choice picks up only the highest order $\Delta^3$ term, making the principal symbol degenerate and hence, strictly speaking, non-elliptic. However, we can plainly see that the system is just a combination of three independent elliptic equations of different orders. Selecting the weights/gradings to be $b_j = 0$ and $a_1 = 6$, $a_2 = 4$ and $a_3 = 2$ allows the weighted principal symbol to pick up all the components (in this example there are no subleading terms) and become non-degenerate. The ADN definition of ellipticity simply generalizes this idea. The operator $D$ is _ADN elliptic_ if there exists a choice of weights/gradings such that its weighted principal symbol (leading homogeneous term) is invertible for $\xi\ne 0$.

Finally, to the question whether there exists some automatic way of choosing the weights, in general I don't know. Certainly, it's possible that an operator is not elliptic with respect to any weights. Perhaps there is a way to decide when an operator can be made elliptic with respect to _some_ choice of weights, but I don't know it of the top of my head. Possibly this question can be answered within commutative algebra and the theory of graded modules.