If $A$ is a finite dimensional algebra over a field, then, by Krull-Schmidt, every finitely generated $A$-module
can be decomposed into a (finite) direct sum of indecomposable submodules
the decomposition is unique up to isomorphism.
Now suppose $A$ is a torsion-free $\mathbb{Z}$-algebra of finite $\mathbb{Z}$-rank.
Q1: Is there a Krull-Schmidt theorem for finitely generated $A$-modules ?
To be more precise: By induction on the $\mathbb{Z}$-rank, it follows that every finitely generated torsion-free module is a direct sum of indecomposables. So 1. holds in this context.
Q2: Do all such decompositions into indecomposables have the same number of summands ?
N.b.: By "torsion-fre" I mean torsion-free as abelian group and the $\mathbb{Z}$-rank is just the rank of the underlying free abelian group.
Generally 2. does not hold, i.e. the Krull-Schmidt Theorem does not hold for f.g. $A$-modules.
Let $A = \mathbb{Z}[\zeta_{23}]$, where $\zeta_{23} \in \mathbb{C}$ is a 23rd root of 1. The projective module group $K_0(\mathbb{Z}[\zeta_{23}])$ is the same as the ideal class group of $\mathbb{Z}[\zeta_{23}]$, which is not the trivial group. Thus, there are two projective (and hence torsion-free) $\mathbb{Z}[\zeta_{23}]$ which are not free, but whose direct sum is free.
This only shows the failure of 2. I do not have an example which shows the answer to Q2 is no.
Perhaps relevant: there is a body of work on torsion-free Krull-Schmidt Theorems for integral group rings. Example with some references: "On the Krull-Schmidt-Azumaya Theorem for Integral Group Rings", by Peter Hindman, Lee Klingler and Charles Odenthal.
Also perhaps relevant, there is a Krull-Schmidt Theorem for algebras over the p-adic integers.