$\mathbb{Z}_n$ is $J$-semisimple iff $n$ is square free $\mathbb Z_n$ is $J$-semisimple if and only if $n$ is square free.

A ring  $R$ is said to be $J$-semisimple if intersection of all maximal ideals of $R$ is $\{0\}$.

If $n$ is square free, then $n = p_ 1p_2 \dotsm p_k$, then $\mathbb{Z}_n \cong \mathbb{Z}_{p_1} \times \mathbb{Z}_{p_2} \times \dotsb \times \mathbb{Z}_{p_k}$. Since $\mathbb Z_{n}$  is a PID every prime ideal ideal is a maximal ideal and hence the maximal ideals will be $\mathbb{Z}_{p_i}$ for each $i$; but the intersection of all $\mathbb{Z}_{p_i}$’s is $\{ 0 \}$ since the $p_i$’s are distinct.
How to do the converse?
 A: Write $n=p_1^{e_1}\dotsm p_r^{e_r}$. Then
$$\mathbf Z/n\mathbf Z\simeq\mathbf Z/p_1^{e_1}\mathbf Z\times\dotsm\times\mathbf Z/p_r^{e_r}\mathbf Z $$
If $n$ is square-free, $\mathbf Z/n\mathbf Z$ is a product of fields, , which are semi-simple rings, hence it is semi-simple.
Conversely, if $\mathbf Z/n\mathbf Z$ is semi-simple, each $\mathbf Z/p_i^{e_i}\mathbf Z$ is a quotient of $\mathbf Z/n\mathbf Z$, hence is semi-simple. As it has only one maximal ideal: $\, p_i\mathbf Z/p_i^{e_i}\mathbf Z$, it must be $0$, i.e. $\,p_i\mathbf Z=p_i^{e_i}\mathbf Z$, which implies $\,e_i=1$. Thus $n$ is square-free.
A: Let's show that if $n = p_1^{e_1} \dotsb p_k^{e_k}$ with $e_i \geq 1$ for every $i \in \{1,\dotsc,k\}$, then the maximal ideals of $\Bbb{Z}_n \simeq \Bbb{Z}_{p_1^{e_1}} \times \dotsb \times \Bbb{Z}_{p_k^{e_k}}$ are precisely those of the form
$$
M_i := \Bbb{Z}_{p_1^{f_1}} \times \dotsb \times \Bbb{Z}_{p_k^{f_k}}
$$
where $f_i = e_i - 1$ for exactly one $i \in \{1,\dotsc,k\}$ and $f_j = e_j$ for $j \neq i$. (1)
Indeed, $\Bbb{Z}_n / M_i \simeq \Bbb{Z}_{p_i}$ is a field. Vice-versa, if $I = \Bbb{Z}_{p_1^{r_1}} \times \dotsb \times \Bbb{Z}_{p_k^{r_k}}$ with $r_i < e_i$ and $r_j < e_j$ for some $i,j \in \{1,\dotsc,k\}$, then $\Bbb{Z}_n / I$ contains a subring isomorphic to $\Bbb{Z}_{p_i} \times \Bbb{Z}_{p_j}$, so it isn't even a domain. Similarly if $r_i \leq e_i - 2$ for some $i$ and $r_j = e_j$ for $j \neq i$.
Now that we have this, can you tell what is $J(\Bbb{Z}_n) = M_1 \cap \dotsb \cap M_k$ if $n$ is squarefree? What if $e_i > 1$ for some $i \in \{1,\dotsc,k\}$, instead?

(1) Note that here $\Bbb{Z}_{p_s^{f_s}}$ denotes the ideal of $\Bbb{Z}_{p_s^{e_s}}$ generated by $p^{e_s - f_s}$ as an additive subgroup. For example, the non-trivial ideals of $\Bbb{Z}_8$ are
$$
\{0,2,4,6\} \simeq \Bbb{Z}_4
\qquad \text{and} \qquad
\{0,4\} \simeq \Bbb{Z}_2
$$
where the isomorphisms are intended as isomorphisms of (additive) groups.
A: Let $n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$ be the decomposition as product of powers of distinct primes; then we know that
$$\def\Z#1{\mathbb{Z}/#1\mathbb{Z}}
\Z{n}\cong
\Z{p_1^{e_1}}\times\Z{p_2^{e_2}}\times\dots\times\Z{p_k^{e_k}}
$$
by the Chinese remainder theorem.
If $n$ is squarefree, then this is a product of fields and it's easy to show a set of maximal ideals having zero intersection.
Suppose, without loss of generality, that $e_1>1$ and consider the element
$$
x=([p_1],[0],\dots,[0])
$$
in the direct product, where brackets mean the residue class in the corresponding ring. Then, for every $y=([y_1],\dots,[y_k])$, $1-xy$ is invertible, because
$$
1-xy=([1-p_1y_1],[1],\dots,[1])
$$
and $[1-p_1y_1]$ is invertible in $\Z{p_1^{e_1}}$ (verify it).
Thus $x$ belongs to every maximal ideal (why?) and so the intersection of the maximal ideals is nonzero.
A: In the finite case as in our case the nil radical is the same as the other radicals. Thus we have no non zero nilpotent elements and so n is the product of distinct primes.
