Edit: Per Tapio's answer, $1 \times 2 \times 3 \times ... \times n$ bricks always suffice, so $s(n)\leq \left(\begin{array}{c} n+1 \\ 2 \end{array}\right)$. Consider the lattice in $\mathbb Z^n$ defined by the equations $x_1 \equiv x_2$ mod $2$, $x_2 \equiv x_3$ mod $3$, ..., $x_{n-1} \equiv x_n$ mod $n$. Place a $1 \times 2 \times ... \times n$ bricks with one corner at each vertex of the lattice. This covers $\mathbb R^n$ because you first round $x_1$ down to the nearest integer, then $x_2$ down to the nearest multiple of two plus $x_1$, .... The second condition is clearly satisfied, since the distance between any two bricks is an integer. We only need to check the first. Equivalently, we check that there cannot be $n+2$ lattice points in a single brick of the same size. Proof: Suppose there were. Then two of them would have to have the same value of $x_n$ mod $n+1$. Since the $x_n$ values of those two points lie in an interval of length $n$, they must be the same. So they have the same value of $x_{n-1}$ mod $n$. By induction, they are identical.
Is this bound sharp? For $n=1$, this is clear. Here is a proof that for $s(2)=3$. n=2$. Form a graph where the faces are bricks, the the edges are the boundaries of bricks, and the vertices are places where two bricks intersect. Suppose that no brick is a hexagon or larger. Then the number of edges in a large reason is no more than$5/2$the number of faces, and the number of vertices is exactly$2/3$the number of edges, so the Euler number is at least$F -5/2(1-2/3)F=F/6$which is$O$of the area of the region, where it should be$O$of the boundary. Or "the graph is somewhere between a cube and a dodecahedron, but nowhere near an infinite plane" 1 Here is a proof that$s(2)=3$. Form a graph where the faces are bricks, the the edges are the boundaries of bricks, and the vertices are places where two bricks intersect. Suppose that no brick is a hexagon or larger. Then the number of edges in a large reason is no more than$5/2$the number of faces, and the number of vertices is exactly$2/3$the number of edges, so the Euler number is at least$F -5/2(1-2/3)F=F/6$which is$O$of the area of the region, where it should be$O$of the boundary. Or "the graph is somewhere between a cube and a dodecahedron, but nowhere near an infinite plane" Therefore, some face has at least 6 edges. Each edge has length and least$1$, since the two vertices can share at most two faces, so the other faces at each vertex are nonadjacent, so have distance at least$1$. Therefore the perimeter of some face is at least$6$. The perimeter of$[0,a]\times [0,b]$is of course$2(a+b)$, so$s(2)\geq 3$. There is an explicit example with$s(T)=3\$, so we are done.