I believe that what you say is true. I'll sketch an argument.

Let f:Z^{n} ---> Z^{2n} be the map of free Z-modules given by the matrices B_{1}, B_{2} put in column (i.e. the direct sum of the morphisms given by B_{1} and B_{2}). Now we rephrase conditions (1) and (2) in a slightly more abstract way:

(1) fails to hold if, and only if there exists p:Z^{2n} ---> Z^{n} such that, together with f, fit in a short exact sequence

0 ---> Z^{n} ---> Z^{2n} ---> Z^{n} ---> 0 (*)

Indeed, the failure of (1) means that any v in Q^{n} such f(v) in Z^{2n} must be integral (i.e. v in Z^{n}). In particular, this implies that f is injective. Moreover, take w in Z^{2n} representing a nonzero torsion element in the cokernel of f. As w represents a torsion element, Nw belongs to the image of f for some big enough positive integer N, so there is v in Z^{n} such that f(v) = Nw. But now f(1/N v) = w, and this means, by the failure of (1), that 1/N v is integral, so w is in the image of f and the cokernel of f has no torsion. As a finitely generated torsion-free Z-module is free, we get an exact sequence like (*) above. This argument can easily be reversed, to show the equivalence between the existence of this exact sequence and the failure of (1).

- (2) holds if, and only if there exists a morphism of Z-modules r:Z
^{2n} ----> Z^{n} such that rf = id.

Let r be represented by a matrix (A_{1},A_{2}). Then gf has matrix A_{1}B_{1} + A_{2}B_{2}, and gf = id if, and only if (2) holds.

Now, the proof of what you asked for is easy. (1) fails if, and only if we can form the exact sequence (*), but such an exact sequence is always split because Z^n is projective, so we can form such exact sequence if, and only if there exists a splitting r:Z^{2n} ----> Z^{n}, which is exactly condition (2).