Let $x=(x_1,\ldots,x_n)$ be a real vector in general position and let's say it is normalized: $\lVert x\rVert=1$.
Let $y$ be a real number that can be arbitrarily large and $\epsilon>0$ can be arbitrarily small.
Consider an approximation $|a_1 \, x_1+\ldots+a_n \, x_n - y| < \epsilon$, where $a_1,\ldots,a_n$ are integers.
CORRECTION: it was kindly pointed out by user2566092 (see below) that the above problem is not well-defined. Fortunately, this is not the one I am actually interested in :-)
The approximation I am looking at is
$ 0 < y - a_1 \, x_1+\ldots+a_n \, x_n < \epsilon $
What can be said about the upper bound on sizes of these integers in terms of $|y|$ and $\epsilon$ ?
Alex--
You're going to need some more conditions on $x$ too. Suppose $x_1$ is very close to $1$ and all other $x_i$ are very very small. Then take $y$ to be $1/2$. Then for small $\epsilon$ the bound on the $a_i$ will depend on how small the $x_i$ are. So for any small $\epsilon$, there is no universal bound on the $a_i$ because it depends on how small the small $x_i$ are, i.e. the bounds go to $\infty$ as the small $x_i$ go to zero.
Even if the $x_i$ are bounded away from zero, you can still run into the same problem. Let's say each $x_i = \delta_i \pm 1/\sqrt{n}$ where some $x_i$ are positive and some are negative and all $\delta_i$ are arbitrarily small. Then put $y = 2\epsilon$ for some some positive $\epsilon$. Then as the $\delta_i$ go to zero, the bounds on the $a_i$ go to $\infty$, even though we are keeping $y$ and $\epsilon$ fixed.