= Solution
The <first-order Sobolev space> is $H^1(\Omega)=\{v\in L^2(\Omega):\partial_1v,\partial_2v\in L^2(\Omega)\}$, with derivatives understood as <weak derivatives> and norm squared $\int_\Omega(v^2+|\nabla v|^2)$.
\b[The printed normal boundary condition must be distinguished from a conormal condition.] Take $K$ symmetric positive definite, as required for the symmetric matrix requested later. <Integration by parts> gives
$$
\int_\Omega\nabla v\cdot K\nabla u-\int_{\partial\Omega}v\,\mathbf n\cdot K\nabla u=\int_\Omega fv.
$$
The natural boundary term contains the <conormal derivative> $\mathbf n\cdot K\nabla u$, not the ordinary <normal derivative> $\partial_nu$. On a smooth boundary, writing $\mathbf t$ for its unit tangent, the literal condition $\partial_nu=0$ leaves the term $b\partial_{\mathbf t}u$, where $b=\mathbf n\cdot K\mathbf t$. Its weak form therefore retains a boundary pairing:
$$
\int_\Omega\nabla v\cdot K\nabla u-\left\langle b\,\partial_{\mathbf t}(\operatorname{Tr}u),\operatorname{Tr}v\right\rangle=\int_\Omega fv.
$$
Here the tangential derivative of the <Sobolev trace> can be interpreted in $H^{-1/2}$ and paired against $H^{1/2}$ on a smooth boundary. This form is generally nonsymmetric; the requested usual symmetric energy construction does not follow from the literal boundary condition.
For an explicit failure of the claimed mean-zero compatibility, take the unit disk, $K=\operatorname{diag}(2,1)$ and $u=(x^2-y^2)[1-(x^2+y^2)/2]$. It has $\partial_nu=0$ on the circle, but
$$
f=-\nabla\cdot K\nabla u=-2+12x^2-6y^2,\qquad \int_\Omega f=-\frac\pi2\ne0.
$$
Thus replacing the printed normal condition by a zero conormal flux is not harmless.
For the intended zero-<conormal derivative> problem, the requested variational and finite element conclusions are as follows. Seek $u\in H^1(\Omega)$ with
$$
\boxed{a(u,v):=\int_\Omega\nabla v\cdot K\nabla u=\ell(v):=\int_\Omega fv\quad(v\in H^1(\Omega)).}
$$
Putting $v=1$ gives $\int_\Omega f=0$. On a connected domain this condition permits a solution unique modulo constants; the <Poincare-Wirtinger inequality> gives coercivity after fixing the mean.
The energy is $J(v)=\tfrac12a(v,v)-\ell(v)$. A minimizer satisfies $\frac d{dt}J(u+tv)|_{t=0}=a(u,v)-\ell(v)=0$. Conversely, a weak solution satisfies
$$
J(v)-J(u)=\frac12a(v-u,v-u)\geq0,
$$
so it is a minimizer. Testing by $u$ gives the minimum value
$$
\boxed{J_{\min}=-\frac12\ell(u)=-\frac12a(u,u).}
$$
This value is independent of the arbitrary additive constant when $\int f=0$. If the compatibility fails, constant shifts make $J$ unbounded below.
Write $u_h=\sum_jc_j\phi_j$. The <Galerkin method> gives
$$
\boxed{Ac=F,\qquad A_{ij}=\int_\Omega\nabla\phi_i\cdot K\nabla\phi_j,\qquad F_i=\int_\Omega f\phi_i.}
$$
Symmetry of $K$ gives $A^T=A$. For $v_h=\sum_jz_j\phi_j$, $z^TAz=\int_\Omega\nabla v_h\cdot K\nabla v_h\geq0$. Equality forces $v_h$ constant by positive definiteness and connectedness. Since the basis sums to one, the coefficient vector $\mathbf1=(1,\ldots,1)^T$ represents that constant, giving
$$
\boxed{\ker A=\operatorname{span}(\mathbf1),\qquad \mathbf1^TF=\int_\Omega f=0.}
$$
Thus the <stiffness matrix> is symmetric positive semidefinite and singular; the displayed compatibility condition is necessary and sufficient for the finite system. Positive definiteness of $K$ controls every gradient direction, preventing additional zero-energy directions and supplying ellipticity. A degenerate coefficient can introduce further kernel modes, while an indefinite one destroys the minimizing-energy argument.
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