= Solution
For the full <modular group> $\Gamma=SL_2(\mathbb Z)$ and <integer> weight $k$, a <modular form> is a <holomorphic function> on the <upper half-plane> satisfying
$$
f\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^k f(\tau)
\quad\text{for all }\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma,
$$
and <holomorphic> at the cusp at infinity. Explicitly, periodicity under $\tau\mapsto\tau+1$ gives a <Fourier expansion of a modular form> in $q=e^{2\pi i\tau}$, and holomorphy at the cusp means $f=\sum_{n\geq0}a_nq^n$. A <cusp form> has $a_0=0$. All cusps are equivalent for this group. The <matrix> $-I$ forces $f=(-1)^kf$, so every odd-weight <modular form> is <zero>.
There is a corresponding qualification to the displayed <Eisenstein series> formula: \b[the Bernoulli-normalized formula is for even <integers> $k\geq4$.] For odd <integers> $k>2$, pairing $(m,n)$ with $(-m,-n)$ gives $G_k=0$, while $B_k=0$, so the printed expression divided by $B_k$ is undefined. We prove the asserted nonzero formula in its even-weight range.
On each <compact set> in the <upper half-plane>, the real-linear map $(m,n)\mapsto m\tau+n$ has a uniformly bounded inverse. Hence $|m\tau+n|\geq C(m^2+n^2)^{1/2}$ there. For $k>2$, lattice convergence gives locally uniform <absolute convergence> of $G_k$, and therefore holomorphy. If $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma$, then
$$
m\gamma\tau+n=\frac{(ma+nc)\tau+(mb+nd)}{c\tau+d}.
$$
The map $(m,n)\mapsto(ma+nc,mb+nd)$ permutes the nonzero <integer> pairs. Reindexing proves the weight-$k$ transformation law.
To compute the <Fourier expansion of a normalized Eisenstein series>, start from the <cotangent partial-fraction expansion>
$$
\pi\cot(\pi z)=\frac1z+\sum_{n\geq1}\left(\frac1{z+n}+\frac1{z-n}\right).
$$
For $\operatorname{Im}z>0$, the <cotangent Fourier expansion> is $\pi\cot(\pi z)=-\pi i-2\pi i\sum_{r\geq1}e^{2\pi irz}$. Differentiating both normally convergent expressions $k-1$ times gives
$$
\sum_{n\in\mathbb Z}(z+n)^{-k}=\frac{(-2\pi i)^k}{(k-1)!}\sum_{r\geq1}r^{k-1}e^{2\pi irz}.
$$
For even $k$, the terms with $m=0$ sum to $2\zeta(k)$, and the terms with $m<0$ duplicate those with $m>0$. Substituting $z=m\tau$ in the last identity and collecting $N=mr$ gives
$$
G_k(\tau)=2\zeta(k)+\frac{2(2\pi i)^k}{(k-1)!}\sum_{N\geq1}\sigma_{k-1}(N)q^N,
\qquad
\sigma_{k-1}(N)=\sum_{d\mid N}d^{k-1}.
$$
The rearrangement is absolutely convergent for every $|q|<1$, since the <divisor function> grows at most polynomially.
For completeness, the normalization follows directly from the definition of the <Bernoulli numbers>. Near <zero>, the <cotangent partial-fraction expansion> gives
$$
\pi\cot(\pi z)=z^{-1}-2\sum_{r\geq1}\zeta(2r)z^{2r-1}.
$$
On the other hand $\pi\cot(\pi z)=\pi i+2\pi i/(e^{2\pi iz}-1)$. Substitute $t=2\pi iz$ in the generating function of the <Bernoulli numbers>. The term $B_1=-1/2$ cancels $\pi i$, and the higher odd terms vanish because $t/(e^t-1)+t/2$ is even. Comparison of <Laurent coefficients> proves the <Euler evaluation of even zeta values>
$$
2\zeta(k)=-\frac{B_k(2\pi i)^k}{k!}\quad(k\text{ even},\ k\geq2).
$$
Thus
$$
\boxed{G_k(\tau)=2\zeta(k)\left(1-\frac{2k}{B_k}\sum_{n\geq1}\sigma_{k-1}(n)q^n\right),\qquad k\geq4\text{ even}.}
$$
This <Fourier series> also proves holomorphy at infinity, completing the <modular form> verification.
For the <Ramanujan congruence modulo 691>, the standard facts we use are that the <modular discriminant> is the weight-twelve <cusp form> $\Delta=q\prod_{n\geq1}(1-q^n)^{24}=\sum_{n\geq1}\tau(n)q^n$, and that $S_{12}=\mathbb C\Delta$. The latter also follows from the <valence formula for the modular group>: a nonzero weight-twelve <modular form> has total weighted <zero> order one, so a <cusp form> whose coefficient of $q$ vanishes would have cusp order at least two and must be <zero>. The product makes the <Ramanujan tau function> integer-valued.
Write $E_k=G_k/(2\zeta(k))$. Using $B_4=-1/30$ and the supplied $B_{12}=-691/2730$, we have
$$
E_4=1+240\sum_{n\geq1}\sigma_3(n)q^n,\qquad
E_{12}=1+\frac{65520}{691}\sum_{n\geq1}\sigma_{11}(n)q^n.
$$
Both $E_{12}$ and $E_4^3$ have weight twelve and constant term one. Their difference is a <cusp form>, so is a multiple of $\Delta$. Its coefficient of $q$ is $65520/691-720=-432000/691$, which determines that multiple:
$$
691E_{12}=691E_4^3-432000\Delta.
$$
The coefficients of $E_4^3$ are <integers>. Comparing the positive-index <Fourier coefficients> and reducing modulo $691$ therefore gives $65520\sigma_{11}(n)\equiv-432000\tau(n)$. Both numerical constants are congruent to $566$ modulo $691$, and $\gcd(566,691)=1$. Cancellation proves
$$
\boxed{\tau(n)\equiv\sigma_{11}(n)\pmod{691}\quad(n\geq1).}
$$
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