Is there a congruence subgroup $\Gamma \leq \mathrm{SL}_2(\Bbb Z)$, an integer $k > 0$ and a non-zero modular form $f \in M_k(\Gamma)$ such that $f$ has only finitely many non-zero Fourier coefficients $a_n(f)$ ?
I recall that those coefficients are defined so that $$f(q) = \sum_{n \geq 0} a_n(f) q^n, q = e^{2 \pi i z / h_\Gamma},$$ where $h_\Gamma \geq 1$ is an integer depending only on $\Gamma$.
I know that it is possible for some $f$ to have infinitely many $a_n(f)=0$ (e.g. $a_p(f)=0$ for every odd prime $p$, by taking $f(z)=g(2z)$ for some $g$, see here), but I don't expect that all but finitely many can be zero. However, I don't know how to prove it.
Just to remove it from the unanswered list: this question has been asked and answered at https://mathoverflow.net/questions/391108.