Maybe the following is useful: first pick arbitrary distinct positive numbers $y\_1,\ldots,y\_n$. Note that $(y\_1,\ldots,y\_n,y\_1,\ldots,y\_n,0)$ is a solution to your system for all $l$; however, the last component is zero, so it doesn't fit your constraints. To remedy this, fix $l=n$ and define $F:\mathbb{R}^{n+1}\to\mathbb{R}^n$ by $$F(a_1,\ldots,a\_n,b)\_k = \sum\_{i=1}^n(a\_i + y\_i)^{k+n-1} - b^{k+n-1} - \sum\_{i=1}^n y\_i^{k+n-1}.$$ Note that $F(0) = 0$ and 
$$\frac{\partial F\_k}{\partial a\_i} = (k+n-1)(a\_i+y\_i)^{k+n-2},$$ so 
$$
\begin{align*}\frac{\partial F\_k}{\partial a} &= 
\begin{bmatrix}n(a\_1 + y\_1)^{n-1} & \cdots & n(a\_n + y\_n)^{n-1} \\\\
\vdots & \ddots & \vdots \\\\
(2n-1)(a\_1 + y\_1)^{2n-2} & \cdots & (2n-1)(a\_n + y\_n)^{2n-2}\end{bmatrix} \\\\
 &= \begin{bmatrix}n & & \\\\ & \ddots & \\\\ & & 2n-1\end{bmatrix}
\begin{bmatrix}1 & \cdots & 1 \\\\
\vdots & \ddots & \vdots \\\\
(a\_1 + y\_1)^{n-1} & \cdots & (a\_n + y\_n)^{n-1}\end{bmatrix} 
 \begin{bmatrix}(a\_1 + y\_1)^{n-1} & & \\\\ & \ddots & \\\\ & & (a\_n + y\_n)^{n-1}\end{bmatrix} \\\\
&= D\_1V(a\_1+y\_1,\ldots,a\_n+y\_n)D\_2,
\end{align*}$$
where $D\_1$ and $D\_2$ are nonsingular diagonal matrices and $V(a\_1+y\_1,\ldots,a\_n+y\_n)$ is a Vandermonde matrix, which is nonsingular for all $(a\_1,\ldots,a\_n)$ sufficiently close to zero since $y\_1,\ldots,y\_n$ are distinct. Thus, by the implicit function theorem, for all $b$ sufficiently close to zero, there is a solution $(y\_1+a\_1,\ldots,y\_n+a\_n,y\_1,\ldots,y\_n,b)$ to your system. It should be easy to derive the bounds you require. Morally, though, taking $n=l$ gives you way too many degrees of freedom; you should be able to get away with much smaller $n$. 

<b>Afterthoughts:</b> The problem with the above is that $b$ might need to be taken arbitrarily close to zero as $l$ increases. Since $n=l$ in this argument, if the points $x\_i$ and $y\_i$ also need to be bounded, then this means all the points can't be bounded away from one another as $l$ grows, and the Vandermonde matrix will become increasingly badly conditioned. (I didn't check that rigorously, but the heuristic argument seems pretty convincing.) I don't know how much any of that depends on how $y\_1,\ldots,y\_n$ are chosen. I've made this a community wiki post in case someone can fix the argument or prove that it doesn't work.