= Solution
Write the integrand, without its common overall factor, as $\mathcal L=\partial_+X\cdot\partial_-X-\psi_1\cdot\partial_-\psi_1-\psi_2\cdot\partial_+\psi_2$. Keep each constant <Grassmann variable> on the left. For the first <supersymmetry> component, varying the bosonic term gives
$$
\varepsilon_1\left(\partial_+\psi_1\cdot\partial_-X+\partial_+X\cdot\partial_-\psi_1\right).
$$
Varying its <fermion> term gives $-\varepsilon_1\partial_+X\cdot\partial_-\psi_1+\varepsilon_1\psi_1\cdot\partial_-\partial_+X$. The plus sign in the second term comes from moving the odd parameter past $\psi_1$. The middle terms cancel and the rest form $\varepsilon_1\partial_+(\psi_1\cdot\partial_-X)$. The second <supersymmetry> component similarly gives $\varepsilon_2\partial_-(\psi_2\cdot\partial_+X)$. Hence, off shell,
$$
\boxed{\delta\mathcal L=\varepsilon_1\partial_+(\psi_1\cdot\partial_-X)+\varepsilon_2\partial_-(\psi_2\cdot\partial_+X).}
$$
This proves the local total-derivative identity for rigid <worldsheet supersymmetry>; the <action> is invariant when the corresponding boundary flux vanishes and the transformation preserves the allowed fields. A constant parameter preserves a periodic <boson> and a periodic <fermion>. In a <Neveu–Schwarz sector>, however, $\delta X=\varepsilon\psi$ is antiperiodic for constant $\varepsilon$ and does not preserve the periodic <boson>. Thus the bulk identity does not supply a global constant supercharge in every spin sector. This is the <spin-structure obstruction to constant worldsheet supersymmetry>; the local superconformal formulation handles the different sectors consistently.
For independent field variations, <integration by parts> gives bulk terms
$$
\delta\mathcal L\big|_{\rm bulk}=-2\delta X\cdot\partial_+\partial_-X-2\delta\psi_1\cdot\partial_-\psi_1-2\delta\psi_2\cdot\partial_+\psi_2.
$$
The <equations of motion> are therefore
$$
\boxed{\partial_+\partial_-X=0,\qquad\partial_-\psi_1=0,\qquad\partial_+\psi_2=0.}
$$
For variations vanishing at the temporal ends, the spatial seam term is proportional to the difference between its values at $\sigma=2\pi$ and $\sigma=0$ of
$$
-2\partial_\sigma X\cdot\delta X+\psi_1\cdot\delta\psi_1-\psi_2\cdot\delta\psi_2.
$$
The bosonic part cancels for <closed string> periodicity, also in a fixed winding sector. For a real <fermion> impose $\psi_a(2\pi)=s_a\psi_a(0)$ and the same relation on its variation. Seam cancellation requires $s_a^2=1$, so the two untwisted possibilities are \b[periodic, $s_a=+1$ (<Ramond sector>), or antiperiodic, $s_a=-1$ (<Neveu–Schwarz sector>)]. The two chiralities of a <closed string> can choose these signs independently, giving R–R, NS–NS, R–NS and NS–R sectors. This discussion assumes the ordinary Lorentz-invariant untwisted <boundary conditions> of the flat theory.
The chiral equations and boundary signs give the following <Fourier series>
$$
\psi_1^\mu(\tau,\sigma)=\sum_{r\in\mathbb Z+\nu_1}\psi_{1r}^\mu e^{-ir(\tau+\sigma)},\qquad\psi_2^\mu(\tau,\sigma)=\sum_{r\in\mathbb Z+\nu_2}\psi_{2r}^\mu e^{-ir(\tau-\sigma)},
$$
where $\nu_a=0$ in R and $\nu_a=1/2$ in NS. Thus integer or half-integer modes follow directly from the <Fourier series> monodromy. With the conventional <string oscillator> normalization their anticommutation relations and reality conditions are
$$
\{\psi_{ar}^\mu,\psi_{bs}^\nu\}=\delta_{ab}\eta^{\mu\nu}\delta_{r+s,0},\qquad (\psi_{ar}^\mu)^\dagger=\psi_{a,-r}^\mu.
$$
For an <open string> the endpoint gluing identifies the two sets of coefficients, leaving one R or NS tower.
In ten-dimensional <light-cone gauge in string theory>, the eight transverse bosonic modes have positive integer level and each NS fermionic creator has a positive half-integer level. For the <unprojected low levels of the open RNS string>, all states through $\alpha'M^2=1$ in the <NS sector> are
$$
\begin{array}{c|c|c|c}
N&\alpha'M^2&\text{oscillator states on }|0;p\rangle&\text{number}\\
0&-1/2&|0;p\rangle&1\\
1/2&0&\psi_{-1/2}^i|0;p\rangle&8\\
1&1/2&\alpha_{-1}^i|0;p\rangle,\quad\psi_{-1/2}^i\psi_{-1/2}^j|0;p\rangle\ (i<j)&8+28=36\\
3/2&1&\psi_{-3/2}^i|0;p\rangle,\quad\alpha_{-1}^i\psi_{-1/2}^j|0;p\rangle,\quad\psi_{-1/2}^i\psi_{-1/2}^j\psi_{-1/2}^k|0;p\rangle\ (i<j<k)&8+64+56=128
\end{array}
$$
The antisymmetric index restrictions are consequences of fermionic anticommutation; the <boson>–<fermion> pair has all 64 choices. The displayed partitions exhaust levels at most $3/2$. Here the NS <normal-ordering constant of a string> is $a=1/2$.
In the <R sector> $a=0$, and the zero modes satisfy $\{\psi_0^i,\psi_0^j\}=\delta^{ij}$. Thus $\gamma^i=\sqrt2\psi_0^i$ generate the <Ramond zero-mode Clifford algebra>. The unprojected ground states $|S;p\rangle$ span a sixteen-dimensional transverse <spinor> space, $8_s\oplus8_c$. All states in the requested range are
$$
\begin{array}{c|c|c|c}
N&\alpha'M^2&\text{states}&\text{number}\\
0&0&|S;p\rangle&16\\
1&1&\alpha_{-1}^i|S;p\rangle,\quad\psi_{-1}^i|S;p\rangle&128+128=256.
\end{array}
$$
There is no intervening half-level, since all nonzero R modes have integer level. Acting with the zero modes only moves within the ground <spinor> space and does not generate further independent copies.
The <GSO projection> keeps odd NS fermion-excitation parity, removing the tachyonic vacuum and the entire $N=1$ NS level. In R it keeps one <eigenvalue> of total <fermion> parity, whose zero-mode part is <spinor> <chirality>. It leaves eight massless <spinor> states. At R level one, the retained families are $\alpha_{-1}^i|S_+\rangle$ and $\psi_{-1}^i|S_-\rangle$, of opposite ground <chirality>, so there are $64+64=128$ states. The retained NS states are <spacetime> <bosons> and the R states <spacetime> <fermions>. Consequently \b[the retained levels have $8_B=8_F$ at mass zero and $128_B=128_F$ at $\alpha'M^2=1$, with no <tachyon>].
Equality at every higher mass level follows from the standard <Jacobi abstruse identity>, or equivalently the equality of the transverse <generating functions>
$$
\frac{\prod_{n\geq1}(1+q^{n-1/2})^8-\prod_{n\geq1}(1-q^{n-1/2})^8}{2q^{1/2}\prod_{n\geq1}(1-q^n)^8}=8\frac{\prod_{n\geq1}(1+q^n)^8}{\prod_{n\geq1}(1-q^n)^8}.
$$
The left side counts the projected NS <bosons>, the right side the projected R <fermions>. Their common coefficients start $8+128q+\cdots$. This consistent truncation yields <spacetime> <supersymmetry>; the bulk <worldsheet> symmetry alone would not remove the <tachyon> or impose the projection.
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