In particular cases (say when $\Lambda$ is a rectangular lattice) the solution is obvious. However, I don't really want to start making assumptions about $\Lambda$ , I want to know if there is an algorithm (preferably something fast, like $poly(n)|\mathbb{Z}^N/\Lambda|$) that works in all cases. I can think of an algorithm that works (I think it is similar to what Will is suggesting), but it requires a number of operations that is exponential in $n$, even if $|\mathbb{Z}^N/\Lambda|$ is polynomial in $n$.

@KConrad: $\det{\Lambda}$ is the determinant of the Gram matrix, it is always positive.

If you are looking for practical approaches to outlier removal, you might like to read the answers to mathoverflow.net/questions/6819/…. If you are interested in a philosophical discussion about when it is appropriate to remove outliers (if it ever is) then it appears that Joel might be interested in such a discussion -).

By conditions, do you mean conditions on the distribution of X (and therefore Y)?

I'll assume you are talking about the 'naive algorithm'. The lattice squares (Voronoi cells) don't have to cover the whole quadrant (see the picture). The 'naive algorithm' just takes all of the lattice points with Voronoi cells that intersect the line $\ell$. As pointed out by fedja, this isn't a fast algorithm, we need to devise something faster. The faster algorithm is likely to look similar to Cassels' algorithm or the 'divided cell' algorithm suggested by Alexey.

@Wadim: In this case the Voronoi cell is just a square, so it is not a problem.

@Alexey: I believe that the divided cell algorithm and Cassel's algorithm are very closely related. I was not aware of Delone's paper, thanks!

@Victor: You are correct, if the intersection point is an integer then this becomes homogeneous Diophantine approximation. @Will: No problem.

I don't disagree. The analogue of Cassels' algorithm needs to be found! It might be that Cassels' algorithm works 'as is'. I would be hugely surprised if there was not an analogue of Cassels' algorithm for this problem, but, at the moment there are other things I should be doing (like writing a thesis :)). If someone else can put it all together then all the brownie points to them.