15. (1) When $n=1$, $x_{1} \geqslant y_{1}>0, \frac{1}{x_{1}} \leqslant \frac{1}{y_{1}}$.

When $n=2$, $x_{1}+x_{2} \geqslant y_{1}+y_{2}, x_{1}-y_{1} \geqslant y_{2}-x_{2}, \frac{1}{y_{1}}-\frac{1}{x_{1}}=\frac{x_{1}-y_{1}}{x_{1} y_{1}} \geqslant \frac{y_{2}-x_{2}}{x_{2} y_{2}}=$ $\frac{1}{x_{2}}-\frac{1}{y_{2}}$, so $\frac{1}{x_{1}}+\frac{1}{x_{2}} \leqslant \frac{1}{y_{1}}+\frac{1}{y_{2}}$.

Assume the proposition holds for $n \leqslant k$, for $n=k+1$, let $y_{i}=x_{i}+a_{i}(i=1,2, \cdots, k+1)$. By the condition $a_{1} \leqslant 0, a_{1}+a_{2} \leqslant 0, \cdots, a_{1}+a_{2}+\cdots+a_{k} \leqslant 0, a_{1}+a_{2}+\cdots+a_{k}+a_{k+1} \leqslant 0$, we have
$$\frac{a_{1}}{x_{1} y_{1}}+\frac{a_{2}}{x_{2} y_{2}}+\cdots+\frac{a_{k}}{x_{k} y_{k}} \leqslant 0$$

Assume $\frac{a_{1}}{x_{1} y_{1}}+\frac{a_{2}}{x_{2} y_{2}}+\cdots+\frac{a_{k}}{x_{k} y_{k}}+\frac{a_{k+1}}{x_{k+1} y_{k+1}}>0$, then we have
$$$$$\text{02 k-1}$$. Then$$a_{k+1}>a_{k}+\left(1+2+\cdots+2^{k-2}\right) \geqslant 2^{k-1}+1+$$$$\left(1+2+\cdots+2^{k-2}\right)=2^{k}\$$
By analogy, then $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}<2$.