Let me show that every such $Z$ has finite length.
First, note that being of the form $\varinjlim X_i$ (with the $X_i$ of finite length) is the same as being countably generated over $A$. So let us start with $Z=\varprojlim Y_j$ (with the $Y_j$ of finite length) and prove that $Z$ cannot be countably generated unless it has finite length.
Endow $A$ and the $Y_j$'s with the discrete topology, and $Z$ with the limit topology. Then $Z$ is a pro-discrete (hence complete metrizable) space, and a topological $A$-module. For each $j$, put $U_j:=\ker(Z\to Y_j)$: these are submodules of $Z$ forming a basis of neighborhoods of zero.
Claim. For a submodule $E$ of $Z$, the following conditions are equivalent, and imply that $E$ is closed:
(1) $E$ is discrete;
(2) $Y_j\cap E=\{0\}$$U_j\cap E=\{0\}$ for large $j$;
(3) $E$ has finite length.
Indeed, $(1)\Leftrightarrow(2)$ is clear. (2) implies that $E\to Y_j$ is injective, hence (3); conversely, the $U_j\cap E$ form a decreasing sequence of submodules of $E$, with zero intersection, so (3) implies (2). Finally, if $E$ is discrete it is a locally closed subgroup of $Z$, hence closed.
Now assume that $Z$ is not of finite length. Then it is not discrete (by the above), i.e. $0$ is not isolated. It follows that every discrete (i.e. finite type) submodule has empty interior in $Z$. By Baire's theorem, so does every countable union of such submodules. In other words, $Z$ is not countably generated.