For each $k$, is there a vector bundle $E$ such that $E\oplus\varepsilon^k$ is trivial but $E\oplus\varepsilon^{k-1}$ is not? 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 $\rho(n+1)-1$.
 A: This is more of an extended comment on ways to approach this problem than an answer.

*

*If we have a vector bundle $E$ on $X$ and have a trivialization $E ⊕ \epsilon^k \to \epsilon^{n+k}$, then this determines a map from $X$ to the Stiefel manifold $V_{k,n+k}$ of $k$-dimensional frames in $(n+k)$-dimensional space: at each point $x$ of $X$, we have an isometric embedding $\Bbb R^k \to \Bbb R^{n+k}$ whose orthogonal complement is identified with the fiber $E_x$.


*As a result, the Stiefel manifold $V_{k,n+k}$ is a universal example. The vector bundle $E$ is pulled back from the $n$-dimensional vector bundle $E_{univ}$ on $V_{k,n+k}$. Therefore, $E_{univ} \oplus \epsilon^{k-1}$ is trivial if and only if every $n$-dimensional vector bundle $E$ such that $E \oplus \epsilon^k$ is trivial also has that $E \oplus \epsilon^{k-1}$ is trivial.


*We can also recast this in terms of homotopy theory. The space $V_{k,n+k}$ is the homotopy fiber of the map $BO(n) \to BO(n+k)$ of classifying spaces for vector bundles. We've reduced this to asking if the map $V_{k,n+k} \to BO(n) \to BO(n+k-1)$ is homotopic to the constant map. Moreover, that map factors through the composite $V_{k,n+k} \to S^{n+k-1} \to BO(n+k-1)$, the first map sending a $k$-frame $(v_1,\dots,v_k)$ of ortogonal vectors in $\Bbb R^{n+k}$ to the first element $v_1 \in S^{n+k-1}$, and the second map classifying the tangent bundle of $S^{n+k-1}$.


*In terms of the refined question, we also need to assure that the map $V_{k,n+k} \to BO(n)$ does not lift to a map $V_{k,n+k} \to BO(n-1)$. For the same reason as above, that's equivalent to a cross-section of the map $V_{k+1,n+k} \to V_{k,n+k}$ of Stiefel manifolds. Cross-sections like that almost never exist (see this question).

Now that this is re-expressed in terms of asking if the map $V_{k,n+k} \to BO(n+k-1)$ is homotopic to a trivial map, we could test it by asking more basic questions about homotopy groups. So I'd like to explore that a little. (I apologize, this may not be particularly satisfying: I'm going to start with this geometric problem, and said that we could potentially solve it with hard spectral sequence calculation. Such is the danger with homotopy theorists!)

*

*We can ask: is the map $\pi_*(V_{k,n+k}) \to \pi_*(BO(n+k-1))$ trivial?  If this map is nontrivial, then $E_{univ} \oplus \epsilon^{k-1}$ is not trivial.


*The fibration sequence $V_{k,n+k} \to BO(n) \to BO(n+k)$
gives a long exact sequence on homotopy groups,
$$
\dots \to \pi_* V_{k,n+k} \to \pi_* BO(n) \to \pi_* BO(n+k) \to \pi_{*-1} V_{k,n+1}.
$$
This identifies the image of $\pi_*(V_{k,n+k})$ in $\pi_* BO(n)$ with the collection of elements that map to zero in $\pi_* BO(n+k)$.


*This allows us to rephrase the questions about homotopy groups without needing to reference the Stiefel manifold at all, and ask the equivalent questions: Does there exist an element in $\pi_* BO(n)$ that maps to zero in $\pi_* BO(n+k)$, and which does not map to zero in $\pi_* BO(n+k-1)$?


*Next, let's take a look at what's actually going on with the homotopy groups. In the sequence of spaces $BO(1) \to BO(2) \to BO(3) \to \dots$, each participates in a fibration sequence $S^{m-1} \to BO(m-1) \to BO(m)$, inducing a long exact sequence on homotopy groups:
$$
\dots \to \pi_* S^{m-1} \to \pi_* BO(m-1) \to \pi_* BO(m) \to \pi_{*-1} S^{m-1} \to \dots
$$
Putting all of these long exact sequences together gives a big spectral sequence that starts with the homotopy groups of spheres and converges to the homotopy groups of $\bigcup BO(m) = BO$:
$$
E^1_{p,q} = \pi_{p+q-1} S^p \Rightarrow \pi_{p+q} BO
$$
This spectral sequence is gigantic and extraordinarily messy, even though the thing it is converging to is known by Bott periodicity. It has been heavily studied previously (especially by Mahowald, who related it to the Atiyah-Hirzebruch spectral sequence for the stable homotopy groups of $\Bbb{RP}^\infty$). But our question about homotopy groups is equivalently formulated in terms of this spectral sequence. A class in $\pi_* BO(n)$ whose image is nonzero in $\pi_* BO(n+k-1)$ but zero in $\pi_* BO(n+k)$ will give rise to a differential in this spectral sequence that is a $d_k$ or higher, from the line $p=n+k-1$.
The upshot is this. Take this spectral sequence, starting with all of the homotopy groups of spheres and converging to the homotopy groups of $BO$. If we have arbitrarily long differentials in this spectral sequence, then for any $k$ there is a vector bundle of the type you are asking for.
