Let $\alpha_1, \alpha_2, \dots$ be an infinite sequence of real numbers such that any finite subset is linearly independent over $\mathbb{Q}$. Let $f(N)$ be the number of tuples $(m_1, \dots, m_N)$ with $m_i \in \mathbb{Z}$, $|m_i| \leq N$. for which $$\left|\sum_{i=1}^{N} m_i \alpha_i \right|< \frac{1}{N^{100}}.$$

(where 100 could be your favorite constant, or maybe something sublinear if that's not possible). Is there any known way to put an asymptotic upper bound on $f(N)$ better than the trivial one of $(2N+1)^N$?

For concreteness (or if it makes the question possible to answer!) you could take the $\alpha_i$ to be the square roots of square free integers (ordered however you like, although I do have a particular ordering in mind). If that doesn't work, any effectively computable $\alpha_i$ would be welcome.

It seems that one *might* be able to say something of the sort using Evertse's quantitative subspace theorem but it isn't really clear to me how. Any pointers (or even conjectures) here would be greatly appreciated!

**EDIT:**
I just noticed something, though it's not super useful: If one takes $\alpha_1 = 1$, $\alpha_k = \log p_k$ ($p_k$ the $k$th prime), then for large $N$, for each $i$ it cannot be the case that just $m_1$ and $m_i$ are non-zero by a quantitative Baker's theorem. So we have at least $f(N) \leq (2N + 1)^N - (N-1)(2N+1)^2$. This isn't super useful for me since this has the same asymptotics as the trivial bound, but might be helpful to think about.