For which $n$ is there a regular $n$-simplex with vertices in $\mathbb{Z}^n$ (or equivalently, in $\mathbb{Q}^n$)?

Some easy observations:

Such an $n$-simplex exists with vertices in $\mathbb{Z}^{n+1}$: just take the $n+1$ points $(0, \ldots, 0, 1, 0, \ldots, 0)$.

If $n$ is even and $n+1$ is not a perfect square, then no regular $n$-simplex with vertices in $\mathbb{Z}^n$ exists: on the one hand, the side length $x$ is the square root of an integer, and thus the volume, namely $\frac{\sqrt{n+1}}{n! 2^{n/2}} x^n$, is irrational; on the other hand the volume must be rational by the formula using determinants.

For $n=3$ a regular tetrahedron with vertices in $\mathbb{Z}^3$ does exist: $\{(0,0,0),(1,1,0),(1,0,1),(0,1,1)\}$.

If there exists a Hadamard matrix of order $n+1$, we can normalize it so the first column is all ones and then the remaining $n$ columns give the coordinates of the $n+1$ vertices of a regular $n$-simplex with vertices in $\{1,-1\}^n$.