As far as I understand the question, we need to consider matrices $M=(m_{ij})$ with entries in $[0,1]$ (the integer part does not matter), look at the vectors of $Mq$ modulo $1$ again with $\|q\|_\infty\le N$ (the exact choice of the norm is not important because it merely changes the constant in the final answer) and to show that one of them is $r$-close (again, in the $\ell^\infty$ norm) to a given vector $y$ with $r$ about $N^{\varepsilon}N^{-n/m}$.

Pick up a smooth function $\psi(t)$ supported on $[-1,1]$ and infinitely smooth there such that $\int\psi=1$. Put $\Psi(x)=r^{-m}\prod_{i=1}^m\psi(x_i/r)$ and consider the periodic function $\Phi(x)=\sum_{p\in\mathbb Z^m}\Psi(x-p-y)$. Then we need to show that there exists $q$ with $\Phi(Mq)\ne 0$ unless $M$ belongs to some exceptional set of matrices of measure $N^{-\delta}$, say (then, considering $N=2^k$ and using Borel-Cantelli, we'll arrive at the statement equivalent to the desired one).

Decompose $\Phi(x)$ into its Fourier series $\sum_{\ell\in\mathbb Z^m}c_\ell e(\ell\cdot x)$ where, as usual, $e(t)=e^{2\pi i t}$ and notice that all $|c_\ell|\le 1$, $c_0=1$, and $c_\ell$ are extremely small and decay fast for $\|\ell\|_\infty>r^{-\beta}$ with any fixed $\beta>1$ (the exact condition we need is that the sum of $|c_\ell|$ with $\ell$ in that range is much less than $1$). Now just consider the sum $\sum_{q:\|q\|_\infty\le N}\Phi(Mq)$ and treat each exponent separately. When $\ell=0$, we will get $N^n$. When $\|\ell\|_\infty>r^{-\beta}$, we will get something much smaller even if we add all those terms up. It remains to investigate the sum
$$
\sum_{\ell:0<\|\ell\|_{\infty}\le r^{-\beta}}\left|\sum_q e(\ell\cdot Mq)\right|\,.
$$
To this end, we'll just estimate its average over $M$. Passing $M$ to $\ell$,
we and using the fact that $\left|\sum_q e(z\cdot q)\right|\le C\prod_{j=1}^n\min(N,\|z\|^{-1})$, where $\|z\|$ is the distance from $z$ to the nearest integer, we immediately get the bound $C\log^n N$ for the average of each term in our sum. Thus, the total average is at most $Cr^{-\beta m}\log^n N$, so outside a set of matrices of measure $N^{-\delta}$, we have the whole sum not exceeding $CN^\delta r^{-\beta m}\log^n N$, which is still below $N^n$ if we have $r=N^{\varepsilon}N^{-\frac nm}$ and choose $\beta$ and $\delta$ sufficiently close to $1$ and $0$ respectively depending on $\varepsilon$.