= Solution
Assume $\kappa,\omega>0$ and impose upward <group velocity> in $z>0$. The two positive-frequency components then have wave vectors $\mathbf k_1=(\kappa,m)$ and $\mathbf k_2=(-\kappa,m)$ with $m<0$. Equal lengths are automatic, and orthogonality gives $m^2=\kappa^2$. Hence
$$
\boxed{m=-\kappa,\qquad \omega=N/\sqrt2.}
$$
The given orthogonal-wave condition is therefore possible only at this forcing frequency. The upward <radiation condition> excludes the $m=+\kappa$ branch.
Use <superposition> with vertical amplitude $W$ in each component and <internal-wave polarization>:
$$
\begin{aligned}
w_1&=W\cos(\kappa x-\kappa z-\omega t)+W\cos(-\kappa x-\kappa z-\omega t)\\
&=2W\cos(\kappa x)\cos(\kappa z+\omega t),\\
u_1&=W\cos(\kappa x-\kappa z-\omega t)-W\cos(-\kappa x-\kappa z-\omega t)\\
&=2W\sin(\kappa x)\sin(\kappa z+\omega t).
\end{aligned}
$$
The two sets of crests have slopes $+1$ and $-1$; their energy propagates upward and outward at angles of $45^\circ$, while their phases travel downward. Their interference makes fixed horizontal bands of vertical-velocity nodes, at $\cos\kappa x=0$, and horizontal-velocity nodes, at $\sin\kappa x=0$.
Put $A=\kappa x$, $B=\kappa z+\omega t$. The first-order displacement relative to mean parcel labels is
$$
\xi_1=-\frac{2W}{\omega}\sin A\cos B,\qquad
\zeta_1=\frac{2W}{\omega}\cos A\sin B.
$$
The horizontal displacement correction vanishes identically, while
$$
\xi_1w_{1x}+\zeta_1w_{1z}
=\frac{4\kappa W^2}{\omega}\left(\sin^2A\cos^2B-\cos^2A\sin^2B\right).
$$
Consequently the formal <cross-wave Stokes drift> is
$$
\boxed{\mathbf u_s=\left(0,-\frac{2\kappa W^2}{\omega}\cos(2\kappa x)\right).}
$$
It has alternating downward and upward columns, no depth decay in this ideal infinite plane-wave field, and zero horizontal-period-averaged vertical transport. Each wave alone has zero drift; the interference terms create this result.
This calculation is a kinematic second-order consequence of the linear field, not a complete prediction of mean transport. In particular, an impermeable oscillating lower boundary cannot have that nonzero mean normal parcel velocity. Its first-order displacement is $\zeta_b=(2W/\omega)\cos(\kappa x)\sin\omega t$. Expanding the exact <kinematic boundary condition> on $z=\zeta_b$ gives
$$
w_2(x,0,t)=u_1(x,0,t)\,\partial_x\zeta_b-\zeta_b\,\partial_z w_1(x,0,t),
\qquad
\langle w_2(x,0,t)\rangle=\frac{2\kappa W^2}{\omega}\cos(2\kappa x).
$$
This <Eulerian mean velocity> cancels the computed normal <Stokes drift> at the material boundary. Away from it, the second-order mean-flow and buoyancy equations must also be satisfied. Thus the alternating drift is a valid formal cross-wave effect, but treating it alone as persistent transport with zero Eulerian mean would be physically incomplete. A prescribed permeable boundary would require its own second-order flux specification instead.
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