Yes.

You can think of the odd sphere [math]S^{2n-1}[/math] as living in the complex n-space [math]\mathbb{C}^n[/math], or as living in the real 2n-space [math]\mathbb{R}^{2n}[/math]. You can think of each point [math]x \in S^{2n-1}[/math] as a vector [math](x_1, \dots, x_n)[/math] of complex numbers, or as a vector [math](a_1,b_1,\dots,a_n,b_n)[/math] of real numbers.

Now, note that [math]i \cdot x[/math], where [math]i = \sqrt{-1}[/math] is the imaginary unit, and where the dot denotes scalar multiplication, is perpendicular to [math]x[/math].

Indeed, [math]i \cdot x[/math] as a real vector is [math](-b_1,a_1,\dots, -b_n,a_n)[/math]. So [math] (i \cdot x) \cdot x[/math] (the second dot is the standard dot product in [math]\mathbb{R}^{2n}[/math]) is easily seen to be 0, and hence [math]i \cdot x [/math] and [math]x[/math] are perpendicular.

In other words, [math] i\cdot x[/math] can be thought of as a tangent vector to the point [math]x[/math]. Since each [math]x \in S^{2n-1} \subset \mathbb{C}^n[/math] is nonzero, so too is [math]i\cdot x[/math]. So this is a nonvanishing vector field on the odd sphere [math]S^{2n-1}[/math].