= Solution
Use the mostly-plus <Minkowski metric>, with the last spatial coordinate singled out:
$$
\boxed{x^+=\frac{x^{D-1}+x^0}{\sqrt2},\qquad x^-=\frac{x^{D-1}-x^0}{\sqrt2},\qquad x^I=x^I\quad(I=1,\ldots,D-2).}
$$
Then $2dx^+dx^-=(dx^{D-1})^2-(dx^0)^2$, as required. In these <light-cone coordinates>, $\partial^+=\partial_-$, $\partial^-=\partial_+$ and $\square_D=2\partial_+\partial_-+\partial_I\partial_I$.
Introduce the <Kalb-Ramond field strength>
$$
H_{pmn}=\partial_pB_{mn}+\partial_mB_{np}+\partial_nB_{pm}.
$$
The field equation is $\partial^pH_{pmn}=0$. Under $\delta B_{mn}=\partial_m\alpha_n-\partial_n\alpha_m$, every second-derivative contribution to $\delta H$ cancels by commutation of derivatives. Thus $H$ and its equation are <gauge-invariant>. This is the differential-form identity $d^2=0$ for the <two-form gauge field> $B$.
To impose <light-cone gauge for a two-form>, first choose $\alpha_-=0$ and
$$
\alpha_m=-\partial_-^{-1}B_{-m}\quad(m=+,I).
$$
This sets $B_{-m}$ to zero. A residual transformation preserving the gauge obeys $\partial_-\alpha_m=\partial_m\alpha_-$. With $\chi=\partial_-^{-1}\alpha_-$, this implies $\alpha_m=\partial_m\chi$, including $m=-$. Such a parameter changes $B$ by zero. Therefore \b[no nontrivial gauge transformation of $B$ remains] in the sector where $\partial_-$ is invertible. There is still a redundant description of the gauge parameter itself, $\alpha\mapsto\alpha+d\chi$; the excluded $\partial_-$ zero modes would require separate treatment.
The $(-,n)$ field equations now say $\partial_-\partial^pB_{np}=0$, hence $\partial^pB_{np}=0$. For $n=I$ this gives
$$
-\partial_-B_{+I}+\partial^JB_{IJ}=0,
\qquad\boxed{B_{+I}=\partial_-^{-1}\partial^JB_{IJ}.}
$$
For $n=+$, $\partial^IB_{+I}=0$ follows automatically from antisymmetry of $B_{IJ}$. Therefore the independent components and their equation are
$$
\boxed{B_{IJ}=-B_{JI},\qquad\frac{(D-2)(D-3)}2\text{ independent polarizations},\qquad\square_DB_{IJ}=0.}
$$
The remaining equations follow from this wave equation and the reconstructed longitudinal components.
At the first massless level of the <closed bosonic string>, states $\alpha_{-1}^I\widetilde\alpha_{-1}^J|p\rangle$ carry a product of two transverse vector polarizations. Its symmetric traceless, antisymmetric and trace parts are respectively the <graviton>, the <Kalb–Ramond field> and the <dilaton>. The antisymmetric part has exactly the polarization count just found.
The <gauge-invariant string coupling to a two-form> uses the pullback of the background <Kalb–Ramond field> to the <string worldsheet>:
$$
\boxed{I_B=q\int_\Sigma X^*B=\frac q2\int d^2\sigma\,\epsilon^{ab}B_{mn}(X)\partial_aX^m\partial_bX^n.}
$$
Its variation is $q\int_\Sigma d(X^*\alpha)=q\int_{\partial\Sigma}X^*\alpha$, by <Stokes theorem>. It vanishes on a closed <worldsheet>. For a cylindrical propagation surface it is only an initial/final boundary term, handled by fixed-boundary gauge parameters or the corresponding transformation of external states. Thus the closed string is naturally charged under a <two-form gauge field>.
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