A vector bundle $E$ is stably trivial if $E\oplus\varepsilon^k$ is trivial for some $k \geq 0$; here $\varepsilon^k$ denotes the trivial bundle of rank $k$. For such a bundle, let $s(E)$ denote the smallest value of $k$ such that $E\oplus\varepsilon^k$ is trivial; note that $s(E) = 0$ if and only if $E$ is trivial.
Question: Does every value of $k$ arise as $s(E)$ for some stably trivial bundle $E$?
For $k = 1$, the tangent bundle of a non-parallelisable sphere provides an example, e.g. $s(TS^2) = 1$. In fact, the tangent bundle of any stably parallelisable manifold $M$ which is not parallelisable satisfies $s(TM) = 1$, see this answer.
Some other values of $k$ can be obtained using tangent bundles of spheres as follows. Consider $TS^n$ where $n$ is odd, $n \neq 1, 3, 7$ - note that $TS^n$ is stably trivial but not trivial. As $n$ is odd, it admits a nowhere-zero section. Adams showed that the maximum number of linearly independent sections is $\rho(n+1) - 1$ where $\rho(n+1)$ is the $(n+1)^{\text{st}}$ Radon-Hurwitz number: if $n + 1 = 2^{4a+b}c$ where $a \geq 0$, $0 \leq b \leq 3$, and $c$ is odd, then $\rho(n+1) = 8a + 2^b$. Therefore there is a non-trivial vector bundle $E$ such that $E\oplus\varepsilon^{\rho(n+1)-1} \cong TS^n$ and hence $E\oplus\varepsilon^{\rho(n+1)} \cong TS^n\oplus\varepsilon^1$ is trivial, so $s(E) = \rho(n+1)$. Every $k \equiv 0, 1, 2, 4 \bmod 8$ arises as $\rho(n+1)$ for some odd $n$ with $n \neq 1, 3, 7$.
For the remaining values of $k$ (i.e. $k \equiv 3, 5, 6, 7 \bmod 8$), one can find a bundle $E'$ with $s(E') = k$ in the following way. First choose $a$ such that $k < 8a$. As $\rho(16^a) = \rho(2^{4a}) = 8a$, by the above there is a bundle $E \to S^N$ with $N = 16^a-1$ such that $s(E) = \rho(16^a) = 8a$. As $E\oplus\varepsilon^{8a} = (E\oplus\varepsilon^{8a-k})\oplus\varepsilon^k$, we see that $s(E\oplus\varepsilon^{8a-k}) = k$.
Answer: Yes.
For $k \equiv 0, 1, 2, 4 \bmod 8$, the examples constructed above do not admit a nowhere-zero section, while the examples for $k \equiv 3, 5, 6, 7 \bmod 8$ certainly do. This leads to the following refinement of the original question.
Refined Question: Does every value of $k$ arise as $s(E)$ for some stably trivial bundle $E$ which does not admit a nowhere-zero section?
By the above, we only have the cases $k \equiv 3, 5, 6, 7 \bmod 8$ left to deal with. In particular, the first unknown case is $k = 3$.
One might hope we can modify the construction above by replacing $TS^n$ with $TM$ for some stably parallelisable manifold $M$. However, a theorem of Bredon and Kosinski states that if $M$ is a $n$-dimensional stably parallelisable manifold, then either $M$ is parallelisable or the maximum number of linearly independent vector fields on $M$ is the same as that of $S^n$, namely $\rho(n+1)-1$.