In answer to Pete L. Clark's question Must a ring which admits a Euclidean quadratic form be Euclidean? on Euclidean quadratic forms, I also gave an example in six or fewer variables, repeated below. Pete's Euclidean property (in the case of positive definite integral quadratic forms) is simply that for any point $\vec x \in \mathbf Q^n$ but $\vec x \notin \mathbf Z^n,$ we require that there be at least one $\vec y \in \mathbf Z^n$ such that $$ q(\vec x - \vec y) < 1. $$

The question on the example with 7 variables was a big success, see Verifying an example in the Geometry of Numbers and Quadratic Forms Could some kind soul please verify the example(s) below. Note how very symmetric this one is, I have little doubt that the "worst" point(s) must occur on the main diagonal $x_1 = x_2 = \cdots x_n.$ Indeed, I think that for any point the orthogonal projection onto the main diagonal has worse "Euclidean" value.

For six or fewer variables we can use one of the easiest constructions, include all mixed terms so that the Gram matrix becomes $$ P_6 \; \; = \; \; \left( \begin{array}{cccccc} 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\ \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\ \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} \\\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} \\\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 \end{array} \right) . $$ Then the worst $\vec x$ is either $$ \vec x = \left( \frac{3}{7}, \frac{3}{7}, \frac{3}{7} , \frac{3}{7}, \frac{3}{7}, \frac{3}{7} \right) $$ or $$ \vec x = \left( \frac{4}{7}, \frac{4}{7}, \frac{4}{7} , \frac{4}{7}, \frac{4}{7}, \frac{4}{7} \right) $$ with ``Euclidean minimum'' $\frac{6}{7}.$

This construction is much easier to figure out. In dimension $ n$ we have determinant $\frac{n +1}{2^n}$ and characteristic polynomial $$ \left( \frac{1}{2^n} \right) \left(2 x - (n+1) \right) \left(2 x - 1 \right)^{n-1}. $$ For even $n $ the worst $\vec x$ has either all entries $\frac{n}{2(n + 1)}$ or $\frac{n + 2}{2(n + 1)}$ with a Euclidean minimum of $\frac{n^2 + 2 n}{8 (n+1)}.$ For odd $n $ the worst $\vec x$ has all entries $\frac{1}{2}$ with a Euclidean minimum of $\frac{n+1}{8}.$

The formulas $\frac{n^2 + 2 n}{8 (n+1)}$ and $\frac{n+1}{8}$ show that this recipe fails (just barely) for $n=7$ and more obviously for larger $n.$ I don't believe there are any examples with $n \geq 9$ and I have my doubts that there can be any for $n=8.$ I did try half of Gosset's root lattice for $E_8,$ see http://en.wikipedia.org/wiki/E8_lattice but it does not seem to work to have any of the squared terms with a coefficient other than $1,$ in all likelihood as soon as $n \geq 4.$