The point of this answer is to flesh out my comment above: Any such $X_i$ must be everywhere linearly independent, so they only exist if the manifold $M$ is parallelizable. Suppose, for the sake of contradiction, that the $X_i$ become linearly dependent at some point $p$. Pass to local coordinates $(x_1, \ldots, x_n)$ with $p=(0,0,\ldots, 0)$ and write $$X_i = \sum_j a_{ij}(x) \frac{\partial}{\partial x_j}.$$ Let $A(x)$ denote the matrix $(A_{ij}(x))$. Our hypothesis is that $A(0)$ is not of full rank.
We compute the principal symbolprincipal symbol of $\sum (\partial/\partial X_i)^2$. We have $$\left( \sum_j a_{ij}(x) \frac{\partial}{\partial x_j} \right)^2 = \sum_{j_1, j_2} a_{i j_1}(x) a_{i j_2}(x) \frac{\partial^2}{(\partial x_{j_1}) (\partial x_{j_2})} + \mbox{first order operators}.$$ So the principal symbol is $$\sum_i \sum_{j_1, j_2} a_{i j_1}(x) a_{i j_2}(x) y_{j_1} y_{j_2} = y^T A^T(x) A(x) y$$ where $y$ is the vector $(y_1, y_2, \ldots, y_n)^T$. If $A$ is not of full rank, neither is this quadratic form.
But the symbol of $\Delta$ is positive definite, a contradiction.
