oVlad wrote: Does there exist a positive integer $n{}$ such that for any real $x{}$ and $y{}$ there exist real numbers $a_1, \ldots , a_n$ satisfying \[x=a_1+\cdots+a_n\text{ and }y=\frac{1}{a_1}+\cdots+\frac{1}{a_n}?\] Answer is $\boxed{n=4}$ Easy to show that $x=y=0$ is impossible for $n=1,3$ while $x=0$ and $y\ne 0$ is impossible for $n=2$ For $n=4$ : Choose any $a_4$ such that both $x_1=x-a_4$ and $y_1=y-\frac 1{a_4}$ are nonzero Choose $a_3$ such that $(x_1-a_3)^2\ge 4\frac{x_1-a_3}{y_1-\frac 1{a_3}}$, which is always possible since when $a_3\to +\infty$, $LHS\sim a_3^2$ while $RHS\sim -\frac {4a_3}{y_1}$ solving then $x_1-a_3=a_1+a_2$ and $y_1-\frac 1{a_3}=\frac 1{a_1}+\frac 1{a_2}$ we get $a_1,a_2$ roots of quadratic $A^2-(x_1-a_3)A+\frac{x_1-a_3}{y_1-\frac 1{a_3}}$, which indeed have roots (from suitable choice of $a_3$ as two lines above)

Detailed explanation on why $n=3$ is impossible (the question doesn't require this, but we can find all satisfactory $n$-s if we want to; also $n=1$ and $n=2$ are very obviously impossible with the values set by @pco): Let $x = y = 0$. Clearly, due to the first equation, $a_3 = - a_1 - a_2$. Now, due to the second equation, we get: $\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{-a_1-a_2} = 0$ $\frac{a_1 + a_2}{a_1a_2} = \frac{1}{a_1 + a_2}$ $(a_1+a_2)^2 = a_1a_2$ $a_1^2+a_1a_2+a_2^2 = 0$ and that can't be true unless $a_1$ or $a_2$ or both are equal to $0$ (otherwise, $a_1^2+a_2^2$ has a greater absolute value than $a_1a_2$), but that can't happen. Also, obviously all $n>4$ also work (just let $a_i=1$ for all $i>4$, and we arrive at the case $n=4$, which @pco showed is possible).

I claim that $n=6$ works. Main Idea: Consider the functions $f(x)=x+\cfrac{1}{x}$ and $f(x)=x-\cfrac{1}{x}$ which has ranges $(-\infty, -2]\cup[2, \infty)$ and $(-\infty, \infty)$ respectively. Then, we let $a_2=\cfrac{1}{a_1}$, $a_4=\cfrac{1}{a_3}$ ,so $a_1+a_2+a_3+a_4=a_1+\cfrac{1}{a_1}+a_3+\cfrac{1}{a_3}$ can be any real number, assume $a_1+\cfrac{1}{a_1}+a_3+\cfrac{1}{a_3}=R$. Also let $a_6=-\cfrac{1}{a_5}$, so that $a_5+a_6=a_5-\cfrac{1}{a_5}$ can be any real number, assume that $a_5-\cfrac{1}{a_5}=S$. Then the equation becomes $$\begin{cases*} x=R+S\\ y=R-S \end{cases*}$$Which indeed has a solution.