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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 $n$.