26. $[\mathbf{1 7}]$ For $1 \leq j \leq 2014$, define
$$
b_{j}=j^{2014} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)
$$
where the product is over all $i \in\{1, \ldots, 2014\}$ except $i=j$. Evaluate
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}} .
$$
Answer: $\frac{1}{2014!^{2014}}$ We perform Lagrange interpolation on the polynomial $P(x)=1$ through the points $1^{2014}, 2^{2014}, \ldots, 2014^{2014}$. We have
$$
1=P(x)=\sum_{j=1}^{2014} \frac{\prod_{i=1, i \neq j}^{2014}\left(x-i^{2014}\right)}{\prod_{i=1, i \neq j}^{2014}\left(j^{2014}-i^{2014}\right)}
$$

Thus,
$$
1=P(0)=\sum_{j=1}^{2014} \frac{\left((-1)^{2013}\right) \frac{2014!^{2014}}{j^{2014}}}{(-1)^{2013} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)},
$$
which equals
$$
2014!^{2014} \sum_{j=1}^{2014} \frac{1}{j^{2014} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)}=2014!^{2014}\left(\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}}\right),
$$
so the desired sum is $\frac{1}{2014!^{2014}}$.