Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

1) We need the following quantitative form of the inverse function theorem. Suppose that  $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C_ 1$, that $\max_ {Y\in B(X, \delta)}\|D^2F(Y)\|\le C_ 2$, and that $C_ 1C_ 2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C_ 1})$.  

2) Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y_ 1,\dots,y_ n)_k=\sum_{j=1}^n y_ j^k$ where $k=1,2,\dots,n$. Take $X=(x_ 1,\dots,x_ n)$ where $x_ j=\frac{n+j}{n}$ for $j=1,\dots,n$.

3) Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.

4) Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_ 1,\dots,c_ n)$ to the vector $p(x_ 1),\dots,p(x_ n)$ consisting of the values of the polynomial $p(x)=\sum_ {k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

5) Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.

6) In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.