= Solution
First fix the photon momentum convention. With $Q^2=-q^2>0$, $\nu=P\cdot q>0$ and $x=Q^2/(2\nu)>0$, an absorbed photon gives final momentum $p_X=P+q$. This is the <physical momentum support of a deep-inelastic hadronic tensor>. The original PDF instead prints $\delta^{(4)}(P-p_X-q)$ while keeping that positive-$x$ convention. Taken literally, it would require
$$
p_X^2=(P-q)^2=M^2-Q^2(1+1/x)<0
$$
in the <Bjorken scaling> limit, leaving no physical final states. The sign is therefore inconsistent with the requested nonzero <deep-inelastic structure functions>. Below, use the physically consistent absorption delta function $\delta^{(4)}(P+q-p_X)$. Alternatively all photon-momentum and scaling-variable conventions would have to be reversed together.
Work in the leading <parton model>: the target is unpolarized; scattering is incoherent from massless collinear <quarks> and <antiquarks>; intrinsic transverse momentum and target-mass effects are neglected. The <parton distribution functions> count constituents and include their colour multiplicity. At this order, neglect strong radiative corrections and the resulting logarithmic scale evolution.
Let a quark carry momentum $p=\xi P$, using a lightlike reference $P$ at leading order, and let $r=p+q$. The spin-averaged electromagnetic trace is
$$
\frac12\operatorname{tr}(\not p\gamma^\mu\not r\gamma^\nu)=2[p^\mu r^\nu+p^\nu r^\mu-g^{\mu\nu}p\cdot r].
$$
Integrating its one-particle final <Lorentz-invariant phase space>, with the $1/(4\pi)$ tensor normalization, supplies $\delta_+(r^2)/2$. Thus the <spin-averaged electromagnetic parton tensor> is
$$
w_f^{\mu\nu}=Q_f^2\delta_+(r^2)[p^\mu r^\nu+p^\nu r^\mu-g^{\mu\nu}p\cdot r].
$$
On physical support, $r^2=0$ gives $\xi=x$ and $p\cdot q=Q^2/2$. Define the transverse tensors
$$
t^{\mu\nu}=-g^{\mu\nu}+q^\mu q^\nu/q^2,\qquad \widetilde P^\mu=P^\mu-(\nu/q^2)q^\mu.
$$
The bracket in $w_f$ becomes
$$
\frac{Q^2}{2}t^{\mu\nu}+2\xi^2\widetilde P^\mu\widetilde P^\nu.
$$
This follows by writing $p=\xi\widetilde P-q/2$ on shell and expanding; it also verifies transversality to $q$.
The correct <parton convolution normalization for a hadronic tensor> is
$$
W^{\mu\nu}=\sum_f\int_0^1\frac{d\xi}{\xi}[q_f(\xi)+\bar q_f(\xi)]w_f^{\mu\nu}(\xi P,q).
$$
The $1/\xi$ converts number weighting into covariant target-state normalization: a parton response per constituent is normalized by $2p^0$, whereas the hadron tensor uses $2P^0$, and $P^0/p^0=1/\xi$. Antiquark charge changes sign but its square and symmetric spin trace are the same.
Use the <parton on-shell delta identity>,
$$
\delta(r^2)=\delta(2\nu(\xi-x))=\frac{\delta(\xi-x)}{2\nu}.
$$
Matching the two tensor coefficients now gives
$$
\boxed{F_1=W_1=\frac12\sum_fQ_f^2[q_f(x)+\bar q_f(x)],\qquad F_2=\nu W_2=x\sum_fQ_f^2[q_f(x)+\bar q_f(x)].}
$$
In particular $F_2=2xF_1$, the <Callan-Gross relation> for spin-one-half partons. Scaling here means the naive leading approximation; full <Quantum chromodynamics> produces scale-dependent distributions and corrections.
Each <quark> contributes $+1$ and each <antiquark> contributes $-1$ to its flavour-vector charge. Sea pairs therefore cancel in the <quark number sum rule>:
$$
\boxed{\int_0^1[q_f(x)-\bar q_f(x)]\,dx=N_f.}
$$
Here $N_f$ is the net flavour number, not the total number including sea pairs. This condition by itself fixes no antiquark distribution.
Under the separately imposed valence-only approximation, $\bar q_f=0$ and only up/down distributions remain. The <proton> has $(N_u,N_d)=(2,1)$ and the <neutron> has $(1,2)$. With $Q_u=2/3$, $Q_d=-1/3$, the <valence-only proton and neutron structure-function moments> are
$$
\boxed{\int_0^1\frac{F_2^p(x)}x\,dx=\frac49(2)+\frac19(1)=1,\qquad \int_0^1\frac{F_2^n(x)}x\,dx=\frac49(1)+\frac19(2)=\frac23.}
$$
No extra factor of three belongs here: colour is already included in the constituent distributions and the flavour counts. These are conditional valence-model moments, not unrestricted asymptotic statements in a theory whose evolution generates a sea.
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