  Let $G$ be a finite group, $\Z G$ the integral group ring of $G$ and $\U(\Z
G)$ the group of units of $\Z G$. The Congruence Subgroup Problem for $\U(\Z
G)$ is the problem of deciding if every subgroup of finite index of $\U(\Z G)$
contains a congruence subgroup, i.e. the kernel of the natural homomorphism
$\U(\Z G) \rightarrow \U(\Z G/m\Z G)$ for some positive integer $m$. The
congruence kernel of $\U(\Z G)$ is the kernel of the natural map from the
completion of $\U(\Z G)$ with respect to the profinite topology to the
completion with respect to the topology defined by the congruence subgroups.
The Congruence Subgroup Problem has a positive solution if and only if the
congruence kernel is trivial. We obtain an approximation to the problem of
classifying the finite groups for which the congruence kernel of $\U(\Z G)$ is
finite. More precisely, we obtain a list $L$ formed by three families of finite
groups and 19 additional groups such that if the congruence kernel of $\U(\Z
G)$ is infinite then $G$ has an epimorphic image isomorphic to one of the
groups of $L$. About the converse of this statement we at least know that if
one of the 19 additional groups in $L$ is isomorphic to an epimorphic image of
$G$ then the congruence kernel of $\U(\Z G)$ is infinite. However, to decide
for the finiteness of the congruence kernel in case $G$ has an epimorphic image
isomorphic to one of the groups in the three families of $L$ one needs to know
if the congruence kernel of the group of units of an order in some specific
division algebras is finite and this seems a difficult problem.
