= Solution
Use target-space signature $(-,+,+,+)$ and <conformal gauge> $g_{ab}=e^{2\omega}\operatorname{diag}(-1,1)$. Dots and primes denote differentiation with respect to $\tau$ and $\sigma$. Varying the <Polyakov action> with respect to the <string embedding map> gives the <wave equation>; varying the <worldsheet metric> sets the <worldsheet stress tensor> to zero. Explicitly,
$$
\boxed{\ddot X^\mu-X^{\mu\prime\prime}=0,\qquad \dot X^2+X'^2=0,\qquad \dot X\cdot X'=0.}
$$
With $\sigma^\pm=\tau\pm\sigma$ and $\partial_\pm=\tfrac12(\partial_\tau\pm\partial_\sigma)$, the two <Virasoro constraints> are $(\partial_+X)^2=(\partial_-X)^2=0$.
A fixed spatial circle would sweep out a cylindrical <string worldsheet>. In static coordinates its <string embedding map> is $X=(t,R\cos\theta,R\sin\theta,0)$ and its <induced worldsheet metric> is $\operatorname{diag}(-1,R^2)$. The radial part of the <Nambu–Goto equations of motion> is proportional to $R^{-2}\partial_\theta^2(R\cos\theta,R\sin\theta)=-(\cos\theta,\sin\theta)/R$, which is nonzero. Thus a nonzero-radius cylinder cannot solve the equations. Motion of labels around this same cylinder is a <worldsheet diffeomorphism> and cannot alter this conclusion.
The familiar <pulsating circular string> makes the obstruction particularly explicit. Put $X^0=\kappa\tau$ and $X^1+iX^2=r(\tau)e^{i\sigma}$. The <wave equation> and <Virasoro constraints> reduce to
$$
\ddot r+r=0,\qquad \dot r^2+r^2=\kappa^2.
$$
A constant $r=R\ne0$ violates the first equation. Even rotating the labels, $X^1+iX^2=Re^{i(\sigma+\Omega\tau)}$, cannot help: the <wave equation> requires $\Omega^2=1$, whereas $\dot X\cdot X'=R^2\Omega$ requires $\Omega=0$.
Nor does a rigid rotation of the circular plane rescue the usual circular ansatz in three spatial dimensions. Write its spatial part as $A(\tau)\cos\sigma+B(\tau)\sin\sigma$, with $A^2=B^2=R^2$ and $A\cdot B=0$. The <wave equation> gives $\ddot A=-A$ and $\ddot B=-B$. The constant norms give $A\cdot\dot A=B\cdot\dot B=0$ and $\dot A^2=\dot B^2=R^2$. The mixed <Virasoro constraint>, together with the derivative of $A\cdot B=0$, gives $\dot A\cdot B=A\cdot\dot B=0$. Differentiating $A\cdot B$ twice gives $\dot A\cdot\dot B=0$. This would require four mutually orthogonal nonzero spatial vectors in $\mathbb R^3$. This obstruction is specific to the stated spatial dimension; a <rigid circular string in four spatial dimensions> has more room.
To express the <Virasoro constraints> in modes, extend the <string oscillators> by
$$
\alpha_0^\mu=\widetilde\alpha_0^\mu=\sqrt{\frac{\alpha'}2}\,p^\mu.
$$
Differentiating the two chiral expansions gives $\partial_-X^\mu=\sqrt{\alpha'/2}\sum_n\alpha_n^\mu e^{-in\sigma^-}$ and the corresponding expression with tildes for $\partial_+X$. Equating every <Fourier coefficient> of their squares to zero yields
$$
L_m=\frac12\sum_{n\in\mathbb Z}\alpha_{m-n}\cdot\alpha_n=0,\qquad \widetilde L_m=\frac12\sum_{n\in\mathbb Z}\widetilde\alpha_{m-n}\cdot\widetilde\alpha_n=0\quad(m\in\mathbb Z).
$$
Reality of the <string embedding map> means $\alpha_{-n}=\alpha_n^*$ and $\widetilde\alpha_{-n}=\widetilde\alpha_n^*$. In particular, define the classical <string level operators> by $N=\sum_{n>0}\alpha_{-n}\cdot\alpha_n$ and $\widetilde N=\sum_{n>0}\widetilde\alpha_{-n}\cdot\widetilde\alpha_n$. The zero-mode <Virasoro constraints> become
$$
0=L_0=\frac{\alpha'}4p^2+N,\qquad 0=\widetilde L_0=\frac{\alpha'}4p^2+\widetilde N.
$$
Consequently the classical <bosonic string mass spectrum> is
$$
\boxed{M^2=-p^2=\frac{4N}{\alpha'}=\frac{4\widetilde N}{\alpha'},\qquad N=\widetilde N.}
$$
The equality of the two levels is <closed-string level matching>. There is no <string intercept> here: that shift comes from quantum <normal ordering>, not from these classical equations.
For the <compact boson>, closed-string periodicity permits a winding number $w\in\mathbb Z$, so $X^3(\sigma+2\pi)-X^3(\sigma)=2\pi wR$. Let $q$ be the momentum in this direction. Its zero-mode expansion and the corresponding chiral momenta are
$$
X^3=x^3+\alpha' q\tau+wR\sigma+\text{oscillators},\qquad p_L^3=q+\frac{wR}{\alpha'},\qquad p_R^3=q-\frac{wR}{\alpha'}.
$$
Thus replace the equal left and right zero modes by $\widetilde\alpha_0^3=\sqrt{\alpha'/2}\,p_L^3$ and $\alpha_0^3=\sqrt{\alpha'/2}\,p_R^3$; the nonzero <string oscillators> remain integer-moded. Classically $q$ is continuous. Quantizing the <compact boson> makes momentum wavefunctions single-valued around its circle, giving $q=n/R$ with $n\in\mathbb Z$.
If $k^a$, $a=0,1,2$, is the noncompact momentum and $M^2=-k^ak_a$, the two zero-mode <Virasoro constraints> give
$$
M^2=\left(q+\frac{wR}{\alpha'}\right)^2+\frac{4\widetilde N}{\alpha'}=\left(q-\frac{wR}{\alpha'}\right)^2+\frac{4N}{\alpha'}.
$$
Averaging and subtracting gives the <momentum and winding modes> and <compact-circle closed-string level matching> formulas
$$
\boxed{M^2=q^2+\frac{w^2R^2}{\alpha'^2}+\frac{2(N+\widetilde N)}{\alpha'},\qquad \widetilde N-N+qwR=0.}
$$
The latter becomes $\widetilde N-N+nw=0$ on imposing quantum momentum quantization. The sign follows from assigning tildes to the left mover $\sigma^+$.
An explicit <winding-supported circular string>, valid classically for every $R>0$, is
$$
\boxed{X^0=2R\tau,\quad X^1=R\cos(\sigma+\tau),\quad X^2=R\sin(\sigma+\tau),\quad X^3=R(\sigma-\tau)\pmod{2\pi R}.}
$$
The noncompact projection is a circle of constant radius. Every coordinate obeys the <wave equation>. Its compact winding is $w=1$, and a direct calculation gives
$$
\dot X^2=-2R^2,\qquad X'^2=2R^2,\qquad \dot X\cdot X'=R^2-R^2=0.
$$
Thus both <Virasoro constraints> hold and the <induced worldsheet metric> is $2R^2\operatorname{diag}(-1,1)$, not a degenerate cylinder. The compact motion cancels the mixed stress that spoiled rotation of labels in flat space. Here $q=-R/\alpha'$, $N=0$ and $\widetilde N=R^2/\alpha'$ satisfy the classical <compact-circle closed-string level matching> condition. Requiring this particular classical momentum to equal an individual quantum eigenvalue would impose an additional radius-dependent condition; that is not required to construct the classical solution.
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