= Solution
Let $a=N+\tfrac12$. The useful identity on the square contour is
$$
|\sin\pi(x+iy)|^2=\sin^2\pi x+\sinh^2\pi y.
$$
On the vertical edges the first term equals one; on the horizontal edges $|y|=a$ and $\sinh\pi a>1$. Thus $|1/\sin\pi z|\le1$ on every edge. Also $|z|\ge a$ and the contour has length $8a$, so the <estimation lemma> gives
$$
\left|\oint\frac{dz}{z^2\sin\pi z}\right|\le\frac{8}{a}\longrightarrow0.
$$
At every nonzero integer $k$, the <residue> is $(-1)^k/(\pi k^2)$. At zero the <Laurent series> is
$$
\frac1{z^2\sin\pi z}=\frac1{\pi z^3}+\frac{\pi}{6z}+O(z),
$$
so the <residue> there is $\pi/6$. The positive orientation and the <residue theorem> therefore give
$$
\oint\frac{dz}{z^2\sin\pi z}=2\pi i\left(\frac\pi6+\frac2\pi\sum_{k=1}^N\frac{(-1)^k}{k^2}\right).
$$
Letting $N\to\infty$ yields \b[the alternating reciprocal-square sum]
$$
\boxed{\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}=\frac{\pi^2}{12}.}
$$
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