= Solution
Vanishing net <charge density> does not require vanishing current. Positive and negative charge carriers can cancel in <charge density> while their oppositely directed motions add to a nonzero current. Charge conservation only requires
$$
\partial_t\rho+\nabla\cdot J=0;
$$
for magnetostatics this becomes $\nabla\cdot J=0$.
Because $\nabla\cdot B=0$, one may introduce a <magnetic vector potential> with
$$
B=\nabla\times A.
$$
It is not unique: $A+\nabla\chi$ gives the same field for any <scalar> $\chi$.
For the stated current,
$$
\nabla\cdot J=0,
$$
so it is consistent with stationary charge conservation. Direct calculation gives
$$
\nabla\times J
=\lambda J_0(\sin\lambda z,\cos\lambda z,0)
=\lambda J.
$$
Thus $J$ is a <Beltrami field>. For $\lambda\ne0$, choose
$$
\boxed{B=\frac{\mu_0}{\lambda}J}.
$$
Then $\nabla\cdot B=0$ and $\nabla\times B=\mu_0J$. A convenient Coulomb-gauge potential is
$$
\boxed{A=\frac{\mu_0}{\lambda^2}J},
$$
since $\nabla\times A=B$ and $\nabla\cdot A=0$. <Gradient> gauge terms may of course be added.
If $\lambda=0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is
$$
\boxed{B=(0,0,-\mu_0J_0x),
\qquad
A=\left(0,-\frac12\mu_0J_0x^2,0\right).}
$$
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