It is sufficient to compute
$$
\int_{-\infty}^\infty \frac{w^n dw}{\mathrm{sinch}(\pi w/2)},
$$
since you can shift by $w'$ and expand the numerator in the original integral. Now, this integral converges very well, and there is no harm in adding a factor $\exp(i\epsilon\omega)$ to the integrand, for any $\epsilon>0$. However, now you can deform the contour into the upper half-plane with no harm for the convergence. In this way you can reduce to sum of the residues at the poles $w=2ik$, $k=1,2,\ldots$
$$
\int_{-\infty}^\infty \frac{w^n e^{i\epsilon w} dw}{\mathrm{sinch}(\pi w/2)}=2\pi i\sum_{k=1}^\infty \mathrm{res}(2ik)=-4\pi \sum_{k=1}^\infty(2ik)^nk(-1)^ke^{-2\epsilon k}=2\pi i^n(-\partial_\epsilon)^{n+1}\frac{1}{e^{2\epsilon}+1}.
$$
Now, we have for Euler polynomials
$$
\frac{2e^{xt}}{e^t+1}=\sum_{k=0}^\infty E_k(x)\frac{t^k}{k!},
$$
so that
$$
\frac{1}{e^{2\epsilon}+1}=\sum_{k=0}^\infty 2^{k-1}E_k(0)\frac{\epsilon^k}{k!},
$$
giving for the integral
$$
\left.\int_{-\infty}^\infty \frac{w^n e^{i\epsilon w} dw}{\mathrm{sinch}(\pi w/2)}\right\vert_{\epsilon=0}=\left.2\pi i^n(-\partial_\epsilon)^{n+1}\frac{1}{e^{2\epsilon}+1}\right\vert_{\epsilon=0}=-2\pi (-2i)^nE_{n+1}(0).
$$
So finally
$$
\int_{-\infty}^\infty \frac{w^n dw}{\mathrm{sinch}(\pi w/2)}=-2\pi (-2i)^nE_{n+1}(0).
$$
This is always real, since both sides vanish for odd $n$.

Note that the Euler polynomials form an Appell sequence, and so have a fundamental property
$$
E_n(x+y)=\sum_{k=0}^n \binom{n}{k}E_k(x)y^{n-k},
$$
so a better statement of the result is that (note the change from $\mathrm{sinch}$ to $\sinh$)
$$
\int_{-\infty}^\infty \frac{(w+w')^n dw}{\mathrm{sinh}(\pi w/2)+i0}=(-2i)^{n+1}E_{n}\left(\frac{iw'}{2}\right),
$$
where $+i0$ in the denominator tells you that your integration contour should go above the pole at $w=0$ in the complex plane. Curiously, this integral representation of Euler polynomials seems to be absent from Wolfram functions site. However, I guess it should be known.