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This is from the beginning of the first section of the paper "Second Chern class and Riemann-Roch for vector bundles on resolutions of surfaces singularities" by J. Wahl.

$C$ is a smooth projective curve over $\mathbb{C}$. $E$ is a vector bundle of degree $e$ and rank $n$ (I believe $n$ should be at least $2$). If $E$ is semi-stable, then $h^0(E) \leq e+n$; otherwise one can find a section of $E$ vanishing at more than $\frac{e}{n}$ points, producing a line subbundle of $E$ of slope $> \frac{e}{n}$.

So I think it means that if $h^0(E) > e+n$, then there is a divisor $D$ of degree $\frac{e}{n}$ such that $h^0(E \otimes \mathcal{O}(-D)) > 0$. I don't know how to show it. It seems that Riemann-Roch is not enough.

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    $\begingroup$ Let $p$ be the smallest integer $>\frac{e}{n} $; then $np-e\leq n$. If $x$ is a point of $C$, $h^0(E(-x))\geq h^0(E)-n$. So for any effective divisor $D$ of degree $p$, $h^0(E(-D))> e+n-np\geq 1$. $\endgroup$
    – abx
    Commented Dec 7, 2020 at 14:40

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Here's a possible way to see this.

One chooses the smallest integer $d$ such that $e-nd<0$. Note that $nd \le e+n$.

Then if $D$ is effective of degree $d$, $h^0(E(-D))=0$ since $E(-D)$ is semistable with negative degree.

Since $h^0(E)\le h^0(E(-D))+nd$ (by using the exact sequence $0\to \mathcal O(-D)\to \mathcal O \to \mathcal O_D\to 0$ tensorized by $E$), the result follows from the inequality $nd\le e+n$ observed above.

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  • $\begingroup$ @abx you beat me to it! $\endgroup$
    – Henri
    Commented Dec 7, 2020 at 14:53
  • $\begingroup$ Ah, I forgot to use the stability condition. I thought it is a general result for any bundles... $\endgroup$
    – WWK
    Commented Dec 7, 2020 at 15:44

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