I have thought of a solution and I am posting it here in case somebody is interested.
Claim: let $\mathcal{F}$ be a coherent sheaf on $\mathbb{P}^N_k$, with $k$ algebraically closed and $N\geq 0$. Then $H^0(\mathcal{F}(n))=0$ for $n\ll 0$ if and only if $\mathcal{F}$ has no associated closed points.
Proof of Claim: $(\implies)$ First, we prove that if $\mathcal{F}$ has an associated closed point, then $H^0(\mathcal{F}(n))\ne 0$ for any $n\in \mathbb{Z}$. Since the associated points of $\mathcal{F}$ are the same as the associated points of $\mathcal{F}(n)$, we can suppose $n=0$. Let $S=S[X_0,\dots,X_N]$ be the coordinate ring of $\mathbb{P}^N$ and let $M$ be a graded $S$-module such that $\mathcal{F}=\widetilde{M}$. Then if $p\in \mathbb{P}^N$ is a closed point associated to $\mathcal{F}$ and $\mathfrak{p}\subseteq S$ is the corresponding homogeneous ideal, then $\mathfrak{p}$ is an associated prime of $M$, meaning that there is an embedding
$$ 0 \longrightarrow S/\mathfrak{p}(n) \longrightarrow M $$
for a certain $n\in \mathbb{Z}$. After sheafification we get a morphism of sheaves
$$ 0 \longrightarrow \kappa(p)(n) \longrightarrow \mathcal{F} $$
and taking global sections it follows that $H^0(\mathcal{F})\supseteq H^0(\kappa(p)(n)) \ne 0$.
$(\impliedby)$ For the converse, suppose that we have a coherent sheaf $\mathcal{F}$ on $\mathbb{P}^N$ such that $H^0(\mathcal{F}(n))\ne 0$ for infinitely many $n<0$, then we want to prove that $\mathcal{F}$ has an associated closed point. Let $M=\sum_{n\in \mathbb{Z}} H^0(\mathcal{F}(n))$: then this is a graded $S$-module such that $M_n$ is of finite dinmension over $k$ for every $n$, and such that, if $m\in M_n$ and $\ell\cdot m =0$ for each $\ell \in S_1$, then $m=0$. Indeed, that is equivalent to saying that the global section $m\in H^0(\mathcal{F}(n))$ vanishes on $\{ \ell \ne 0 \}$ for every $\ell\in S_1$, and since these open subsets cover $\mathbb{P}^n$, it must be that $m=0$. Moreover, the associated primes of $M$ give associated points of $\mathcal{F}$, so that we are done if we can prove the following.
Lemma: Let $S=k[X_0,\dots,X_N]$ and let $M=\sum _{n\in \mathbb{Z}}M_n$ be a graded $S$-module such that:
- every $M_n$ is of finite dimension as a vector space over $k$.
- $M$ is saturated, i.e. if $m\in M_n$ is such that $\ell\cdot m =0$ for every $\ell \in S_1$, then $m=0$.
Then, if $M_n\ne 0$ for infinitely many $n<0$, the module $M$ has the ideal of a closed point in $\mathbb{P}^N=Proj(S)$ as an associated prime.
Proof of Lemma: First we observe that $M_n\ne 0$ for every $n\in \mathbb{Z}$. Indeed, it is enough to prove that if $M_n\ne 0$ then $M_{n+1}\ne 0$ as well. So, let $m\in M_n$ be a nonzero element: then by saturatedness there exists an element $\ell\in S_1$ such that $\ell\cdot m \ne 0$ and then this is a nonzero element in $M_{n+1}$. Now we prove the lemma by induction on $N$:
Basic step: we start with $N=0$: in this case, we know that since $M_n\ne 0$ for every $n$, the sheaf $\widetilde{M}$ on $\mathbb{P}^0$ is nonzero, so that it has an associated point, that must be the unique point of $\mathbb{P}^0$, that in particular is closed.
Induction step: let $N\geq 1$. Now, suppose that we can find an element $\ell \in S_1$ such that the submodule
$$ (0:_M \ell) = \{m\in M \, |\, \ell\cdot m = 0\} $$
is nonzero in infinitely many negative degrees. Then we can regard this as a module over $S/\ell S$ and by induction hypothesis it follows that it must contain a prime associated to a closed point in $\mathbb{P}^{N-1}=Proj(S/\ell S)$. Such a prime is of the form $\mathfrak{p}/\ell S$ for $\mathfrak{p}\subseteq S$ and we have inclusions of $S$-modules
$$ S/\mathfrak{p}(n) = \frac{S/\ell S}{\mathfrak{p}/\ell S}(n) \subseteq (0:_M \ell) \subseteq M $$
and this means that $\mathfrak{p}$ is associated to $M$ and is the prime that we were looking for.
Now, if we cannot find any such $\ell$ this means that for every $\ell \in S_1$ the maps $M_n \overset{\cdot\ell}{\to} M_{n+1}$ are injective for all $n\ll 0$. In particular, it follows that $\dim_k M_n \leq \dim_k M_{n+1}$ for all $n\ll 0$ and since these dimensions are all positive and finite, they must stabilize at a certain point, meaning that $\dim_k M_n = \dim_k M_{n+1}$ for $n\ll 0$. In particular, we can find an $n_0 \in \mathbb{Z}$ such that all the maps $M_n \overset{\cdot X_0}{\to} M_{n+1}$ are isomorphisms if $n\leq n_0$. Suppose now that for every $\ell\in S_1$ the maps $M_n \overset{\cdot\ell}{\to} M_{n+1}$ are isomorphisms for $n\leq n_0$ and, for any such $n$ consider the multiplication map
$$ S_1 \otimes_k M_n \to M_{n+1} $$
this map is injective on both factor separatedly, so that, since $k$ is algebraically closed, by a theorem of Hopf (cfr ACGH-Moduli of Alg Curves I, p. 108) it follows that
$$ \dim_k M_{n+1} \geq \dim_k S_1 + \dim_k M_n -1 = \dim_k M_n + N > \dim_k M_n $$
but this is absurd because we know that both dimensions are the same. Hence, there must be an $\ell \in S_1$ such that $\ell \cdot m = 0$ for a certain nonzero element $m\in M_{\overline{n}}$, with $\overline{n}\leq n_0$. But then we can prove that $(0:_M \ell)_n\ne 0$ for every $n\leq \overline{n}$, that would give a contradiction. To see this, recall that all the maps $M_{n}\overset{\cdot X_0}{\to} M_{n+1}$ are isomorphisms for $n\leq \overline{n}$ so that, for every $m\in M_n$ we can denote by $\frac{m}{X_0^r}$ the unique element in $M_{n-r}$ such that $X_0^r \cdot \frac{m}{X_0^r}=m$. Now observe that $X_0\cdot(\ell\cdot \frac{m}{X_0})=\ell \cdot m =0$ but since multiplication by $X-0$ is injective, it follows that $\ell \cdot \frac{m}{X_0}=0$ and then $(0:_{M}\ell)_{\overline{n}-1}\ne 0$. To show that $(0:_{M}\ell)_{\overline{n}-2}\ne 0$ we observe that $X_0\cdot(\ell\cdot \frac{m}{X_0^2})=\ell \cdot \frac{m}{X_0}=0$ and since multiplication by $X_0$ is injective it must be that $\frac{m}{X_0^2}=0$. Continuing like this one can prove that $(0:_M \ell)_n \ne 0$ for all $n\leq \overline{n}$ and this concludes the proof.
