= Solution
Take $0\leq\sigma<2\pi$. Variation of the conformal-gauge <action>, with periodic variations, gives $(\partial_\tau^2-\partial_\sigma^2)X^\mu=0$. Hence $X^\mu=X_R^\mu(\tau-\sigma)+X_L^\mu(\tau+\sigma)$. In an uncompactified direction, periodicity allows equal linear slopes and independent periodic parts. Expanding the periodic parts in a <Fourier series> yields the <closed-string mode expansion>
$$
X^\mu=x^\mu+\alpha'p^\mu\tau+i\sqrt{\frac{\alpha'}2}\sum_{m\ne0}\frac1m\left(\alpha_m^\mu e^{-im(\tau-\sigma)}+\widetilde\alpha_m^\mu e^{-im(\tau+\sigma)}\right).
$$
Here $\alpha_m$ labels the right-moving sector and $\widetilde\alpha_m$ the left-moving sector; exchanging the names is harmless if done consistently. Integration of the <canonical momentum> density $(2\pi\alpha')^{-1}\dot X^\mu$ over the string gives $p^\mu$, fixing the zero-mode coefficient. Reality gives $\alpha_m^\dagger=\alpha_{-m}$ and the analogous relation for tildes, with $\alpha_0=\widetilde\alpha_0=\sqrt{\alpha'/2}\,p$.
The <closed-string physical-state Virasoro conditions> are
$$
(L_0-1)|\Psi\rangle=(\widetilde L_0-1)|\Psi\rangle=0,\qquad L_m|\Psi\rangle=\widetilde L_m|\Psi\rangle=0\quad(m>0),
$$
where $L_0=\alpha'p^2/4+N$ and $\widetilde L_0=\alpha'p^2/4+\widetilde N$. Thus <closed-string level matching> gives $N=\widetilde N$, and $M^2=-p^2=4(N-1)/\alpha'$. \b[The massless level is $N=\widetilde N=1$.] Its general state is
$$
|\zeta;p\rangle=\zeta_{\mu\nu}\alpha_{-1}^\mu\widetilde\alpha_{-1}^\nu|0;p\rangle.
$$
The <string oscillator> commutators give $[L_m,\alpha_n^\mu]=-n\alpha_{m+n}^\mu$: the two terms obtained by commuting through the quadratic generator coincide and cancel its factor $1/2$. Therefore $L_1$ replaces $\alpha_{-1}^\mu$ by $\alpha_0^\mu$, while $L_{m\geq2}$ produces positive <string oscillators> that annihilate this level. The same holds independently for tildes. The complete polarization conditions are
$$
\boxed{p^2=0,\qquad p^\mu\zeta_{\mu\nu}=0,\qquad p^\nu\zeta_{\mu\nu}=0.}
$$
<Null string states> identify $\zeta_{\mu\nu}$ with $\zeta_{\mu\nu}+p_\mu\lambda_\nu+\widetilde\lambda_\mu p_\nu$, with transverse $\lambda,\widetilde\lambda$. At nonzero <momentum> this leaves $24\cdot24=576$ physical polarizations. Their symmetric traceless, antisymmetric and transverse-trace parts are the <graviton>, <Kalb–Ramond field> and <dilaton>.
The original PDF's condition $\alpha'p^2=4$ is inconsistent with calling the displayed excited states massless. It is the on-shell condition of the <string oscillator> vacuum as a physical <tachyon>. For the level-one state, that value would give $L_0=\widetilde L_0=2$, violating the physical conditions. The ket $|0;p\rangle$ used as the basis for a massless excitation must instead carry the excitation's <momentum> $p^2=0$; by itself it is then not an on-shell physical vacuum state.
For the compact coordinate, $X^1(\tau,\sigma+2\pi)=X^1(\tau,\sigma)+2\pi wR$, $w\in\mathbb Z$. Its <momentum> is $n/R$, $n\in\mathbb Z$, since the centre-of-mass wavefunction is single-valued on the circle. Thus its zero modes become
$$
X^1=x^1+\alpha'\frac nR\tau+wR\sigma+\text{the same periodic oscillator sum},
$$
with
$$
p_R=\frac nR-\frac{wR}{\alpha'},\qquad p_L=\frac nR+\frac{wR}{\alpha'},\qquad \alpha_0^1=\sqrt{\frac{\alpha'}2}p_R,\quad\widetilde\alpha_0^1=\sqrt{\frac{\alpha'}2}p_L.
$$
These are <momentum and winding modes>. Let $k$ be the 25-dimensional noncompact <momentum> and $M^2=-k^2$. The two zero-mode constraints now give
$$
M^2=p_R^2+\frac4{\alpha'}(N-1)=p_L^2+\frac4{\alpha'}(\widetilde N-1).
$$
Their difference and average give, respectively,
$$
\widetilde N-N+nw=0,\qquad M^2=\frac{n^2}{R^2}+\frac{w^2R^2}{\alpha'^2}+\frac2{\alpha'}(N+\widetilde N-2).
$$
This also displays <T-duality> $R\leftrightarrow\alpha'/R$, $n\leftrightarrow w$, with one chiral <momentum> reversed.
At a generic radius, the massless solutions are only $n=w=0$, $N=\widetilde N=1$. Their fields in 25 dimensions are a <graviton> (275 polarizations), antisymmetric two-form (253), <dilaton> (one), two abelian <vectors> from $G_{a1}$ and $B_{a1}$ (23 each), and the radius scalar (one). Hence \b[the generic massless spectrum has $275+253+1+46+1=576$ polarizations], with <gauge group> $U(1)_L\times U(1)_R$. Here generic excludes isolated radii where <exceptional massless ground states of a bosonic circle> occur. Indeed $N=\widetilde N=0$ forces $nw=0$, and becomes massless at $R=|n|\sqrt{\alpha'}/2$ or $R=2\sqrt{\alpha'}/|w|$. These additional thresholds matter in the bosonic theory.
At the <self-dual circle> $R=\sqrt{\alpha'}$, masslessness and level matching read
$$
n^2+w^2+2(N+\widetilde N-2)=0,\qquad\widetilde N-N+nw=0.
$$
Since $N,\widetilde N\geq0$, their sum is at most two. For sum two, $n=w=0$ and level matching gives the ordinary $(1,1)$ family. For sum one the four possibilities are
$$
(N,\widetilde N;n,w)=(1,0;1,1),\ (1,0;-1,-1),\ (0,1;1,-1),\ (0,1;-1,1).
$$
They have one right-moving $\alpha_{-1}$ or one left-moving $\widetilde\alpha_{-1}$ excitation, as their levels specify. Each family has 24 physical polarizations: a 25-dimensional <vector> with 23 polarizations and one compact-oscillator scalar. Thus these families add 96 states.
For sum zero, level matching forces $nw=0$, while the mass equation requires $n^2+w^2=4$. It gives four additional scalar states with charges \b[$(\pm2,0)$ and $(0,\pm2)$], without any <string oscillators>. They must not be omitted. The <massless spectrum at the bosonic self-dual circle> therefore has \b[100 additional polarizations, 676 in total]. The four new <vector> fields enhance the <gauge group> to $SU(2)_L\times SU(2)_R$. Including the original radius scalar, the nine non-dilaton scalars form its $(3,3)$ representation. Equivalently, the full count is $275+253+1+6\cdot23+9=676$.
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