Yes. Let $D$ be a first order elliptic operator from $V$-valued functions to $W$-valued functions. Consider the operator

$$ P = \begin{pmatrix} 0 & D^{\ast} \\ D & 0 \end{pmatrix} $$

acting on the vector bundle $V \oplus W$. The operators $P$ is formally selfadjoint and graded. Let $p(x, \xi)$ be the principal symbol, which is selfadjoint in the sense that $p(x,\xi)^{\ast}=p(x,\xi)$. Consider the square of the symbol $p(x,\xi)^2$, which is selfadjoint and positive definite (since it is the square of a selfadjoint linear map). Thus $p(x,\xi)^2 = q(x,\xi)$, with $q$ quadratic and positive definite in $\xi$. Define a new Riemann metric on $\mathbb{R}^n$ by $|v|^2 := q(x,v^{\sharp})$, when $v \in T_x \mathbb{R}^n$ and $\sharp$ is the musical isomorphism. You find out that the symbol of $P$ defines a Clifford structure on the trivial bundle $V \oplus W$, but with a possibly different metric on $\mathbb{R}^n$. Then $P$ is a Dirac type operator, except that it might not be selfadjoint with respect to the new metric, but at least its symbol is self-adjoint, and so $P$ is selfadjoint up to order $0$ operators.

In particular, the vector space $V \oplus W$ will have the structure of a graded Clifford module for $Cl^n$. Or, $V$ will be an ungraded module over $Cl^{n-1}$. This puts restrictions on $\dim (V)=\dim (W)$. If $n=2m+1$ is odd, then $Cl^{2m} = Mat(2^m, \mathbb{C})$, and $\dim (V)$ is a multiple of $2^m$. If $n=2m$ is even, then $Cl^{2m-1} = Mat(2^{m-1}, \mathbb{C})\oplus Mat(2^{m-1}, \mathbb{C})$, and $\dim (V)$ is a multiple of $2^{m-1}$.

Yes. Let $D$ be a first order elliptic operator from $V$-valued functions to $W$-valued functions. Consider the operator

$$ P = \begin{pmatrix} 0 & D^{\ast} \\ D & 0 \end{pmatrix} $$

acting on the vector bundle $V \oplus W$. The operators $P$ is formally selfadjoint and graded. Let $p(x, \xi)$ be the principal symbol, which is selfadjoint in the sense that $p(x,\xi)^{\ast}=p(x,\xi)$. Consider the square of the symbol $p(x,\xi)^2$, which is selfadjoint and positive definite (since it is the square of a selfadjoint linear map). Thus $p(x,\xi)^2 = q(x,\xi)$, with $q$ quadratic and positive definite in $\xi$. Define a new Riemann metric on $\mathbb{R}^n$ by $|v|^2 := q(x,v^{\sharp})$, when $v \in T_x \mathbb{R}^n$ and $\sharp$ is the musical isomorphism. You find out that the symbol of $P$ defines a Clifford structure on the trivial bundle $V \oplus W$, but with a possibly different metric on $\mathbb{R}^n$ [Here is a gap, see the comments below]. Then $P$ is a Dirac type operator, except that it might not be selfadjoint with respect to the new metric, but at least its symbol is self-adjoint, and so $P$ is selfadjoint up to order $0$ operators.

In particular, the vector space $V \oplus W$ will have the structure of a graded Clifford module for $Cl^n$. Or, $V$ will be an ungraded module over $Cl^{n-1}$. This puts restrictions on $\dim (V)=\dim (W)$. If $n=2m+1$ is odd, then $Cl^{2m} = Mat(2^m, \mathbb{C})$, and $\dim (V)$ is a multiple of $2^m$. If $n=2m$ is even, then $Cl^{2m-1} = Mat(2^{m-1}, \mathbb{C})\oplus Mat(2^{m-1}, \mathbb{C})$, and $\dim (V)$ is a multiple of $2^{m-1}$.