10. (1) By the Cauchy inequality, we have
$$\left(1^{3}+2^{3}+\cdots+n^{3}\right) b_{n+1}^{2} \geqslant\left(1^{3}+2^{3}+\cdots+n^{3}\right)$$
$$\left(\frac{b_{1}^{2}}{1^{3}}+\frac{b_{2}^{2}}{2^{3}}+\cdots+\frac{b_{n}^{2}}{n^{3}}\right)=\left(b_{1}+b_{2}+\cdots+b_{n}\right)^{2}$$

Since $1^{3}+2^{3}+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$, we have
$$\frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}} \geqslant \frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}$$

Therefore,
$$\sum_{n=1}^{k} \frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}} \geqslant \sum_{n=1}^{k}\left(\frac{2}{n}-\frac{2}{n+1}\right)=2\left(1-\frac{1}{k+1}\right)$$

Thus, by taking $k=999$, we get
$$2\left(1-\frac{1}{1000}\right)=\frac{1998}{1000}>\frac{1993}{1000}$$
(2) By taking $k=2001$.