I would like to solve this exercise:
For (Lebesgue) almost every vector $\theta:=(\theta_1,\dots,\theta_d)\in[1,2]^d$, and for every $\varepsilon >0$, there exists a constant $C_{\varepsilon}>0$ such that for every integer $M>0$ and $q=(q_1, \dots,q_d) \in \{-M, \dots, M\}^d \setminus\{0\}$ there holds $$\left\vert \sum_{i=1}^d \theta_iq_i\right\vert \geqslant \frac{C_{\varepsilon}}{M^{d-1+\varepsilon}}.$$
Note that in order to prove this there is a hint, which to use the following theorem (in the following theorem $\|x\|:=d(x,\mathbb{Z}$):
Theorem. Let $q\mapsto \psi (q)$ being a monotonely decreasing function with values in the interval $(0,1/2)$. Then the equation $\|q\theta_i\|<\psi(q) \;\;(1\leqslant j \leqslant d)$ has infinitely many solutions for almost no or almost all $(\theta_1, \dots, \theta_d)$ according as $\sum \psi (q)^d$ converges or diverges.
However I do not see why the exercise follows from such a theorem. We need some multilinear approximation, and also we are restricting to $q \in[-M,M]^d$.
Try: for $d=2$ observe that by the condition $\theta_1, \theta_2 \in [1,2]$ we are left with an equivalent problem asking to prove $|\tilde{\theta}q_1+q_2| \geqslant \frac{C_{\varepsilon}}{M^{1+\varepsilon}}$. Let $\psi (q_1):=q_1^{-1-\varepsilon}$ which is a convergent series then the equation $\|q_1\tilde{\theta}\|<1/q_1^{1+\varepsilon}$, for generic $\tilde{\theta}$ has only finitely many solutions (and even more, we can always choose $\tilde{\beta}$ in a set of full measure so that this distance is $>0$), which means $\|q_1\tilde{\theta}\|> \frac{C}{q_1^{1+\varepsilon}}$. Restricting our attention to $(q_1,q_2) \in [-M,M^2] \setminus\{0\}$ gives the result.
Now I am not able to write a similar proof for dimensions $d \geqslant 3$, and I think there might be simpler arguments! Can someone help me?