**Edit:** According to the comment of Michael Albanese we revise the question.


Assume that $n$ is an odd integer and $M\subset \mathbb{R}^{2n}$ is a compact orientable $n$ dimensional submanifold.


Does there exist a non vanishing smooth vector field $X$ on $M$ with the following property?


$$\forall p\in M\quad \omega(X(p), V\_p)=0,\quad \forall V\_p\in T\_pM$$


where $\omega$ is the standard symplectic structure of $\mathbb{R}^{2n}$.



A special non vanishing vector field on odd dimensional compact manifolds
This is not necessarily true, even locally. Consider $M \subset \mathbb{R}^6$ given by the graph of the function $f : \mathbb{R}\_x^3 \rightarrow \mathbb{R}\_y^3$ with


$$f(x\_1,x\_2,x\_3) = (x\_2x\_3, 0, -x\_1x\_2)$$


Then at a point $p = (\vec{x},f(\vec{x})) \in M \subset \mathbb{R}\_x^3 \times \mathbb{R}\_y^3$, we compute $T\_pM$ is given by the span of the vectors


$$e\_1 = \partial\_{x\_1} -x\_2\partial\_{y\_3}$$
$$e\_2 = \partial\_{x\_2} + x\_3\partial\_{y\_1} - x\_1\partial\_{y\_3}$$
$$e\_3 = \partial\_{x\_3} + x\_2\partial\_{y\_1}$$


Explicitly, we have that in this basis,


$$\omega = \begin{pmatrix} 0 & -x\_3 & -x\_2 \\ x\_3 & 0 & -x\_1 \\ x\_2 & x\_1 & 0 \end{pmatrix}$$


from which it is easy to compute that so long as $p \neq 0$, any $X$ such that $\omega(X,v) = 0$ for all $v \in T\_pM$ (i.e. $X$ is in the nullspace of the above matrix) must satisfy that $$X(p) \in \mathrm{span}\left\langle\begin{pmatrix}x\_1 \\ - x\_2 \\ x\_3\end{pmatrix}\right\rangle$$ By continuity, $X(0)$ would need to be a multiple of $e\_1$ (by looking at $X(p)$ along the axis $x\_2 = x\_3 = 0$) and also a multiple of $e\_3$ (similarly). So there can be no nonzero $X(p)$ in a neighborhood of $0$.


For $M$ in higher dimensions, one can repeat the argument by successively taking $M \times \mathbb{R}^2 \subset \mathbb{R}^{2n} \times \mathbb{R}^4$, where the $\mathbb{R}^2 \subset \mathbb{R}^4$ we choose is symplectic. For a compact $M$, simply choose one which matches this model near $0$.


