This is not a homework problem, although I fear it may turn out to be at that level. For any nonnegative $x\in\mathbb{R}^n$, let $f\_k(x)$ be the sum of the $k$ largest values in $x$, and define $$f(x)=\max\_{k} \frac{f\_k(x)}{\sqrt{k}}$$ over all $k\in\{1,\dots,n\}$. It is easy to see that $f(x)\leq \|x\|\_2$ for all nonnegative $x$. Is the ratio $\|x\|\_2/f(x)$ bounded from above?



The Euclidean norm and $k$ largest elements
I assume that your question concerns only nonnegative elements of $\mathbb{R}^n$, since for instance $f$ vanishes on $(-1,0)\in \mathbb{R}^2$.


Now on $\mathbb{R}\_+^n$, $f$ coincides with $f\circ g$ where $g:\mathbb{R}^n \rightarrow \mathbb{R}\_+^n$ turns every coordinate to its absolute value.
$$f\circ g(x)=\operatorname{max}\_{S\subset\left\lbrace 1, \ldots , n\right\rbrace} \dfrac{\sum\_{i\in S}|x\_i|}{\sqrt{|S|}}\text{ if $x\neq\underline{0}, $ } 0 \text{ otherwise}$$


One easily checks that $f\circ g$ is a norm on $\mathbb{R}^n$, so that the answer is yes since any two norms are equivalent on a finite-dimensional vector space.


However, there is no uniform bound on $n$ as shown by the following example.


Let $x=(x\_1,\ldots, x\_n)\in \mathbb{R}^n$ be defined by $x\_i=\sqrt{i\phantom{.}}-\sqrt{i-1}$. Clearly, $x\_i>x\_{i+1}$ for all $i$, so that 
\begin{eqnarray}
f(x)&=&\max\_{k\in \left\lbrace 1, \ldots , n\right\rbrace} \dfrac{1}{\sqrt{k\phantom{.}}}\sum\_{i=1}^k x\_i \\&=&1
\end{eqnarray}
However, $$\parallel\, x\!\parallel\_2^2 = \sum\_{i=1}^n\left( \sqrt{i\phantom{.}}-\sqrt{i-1}\right) ^2$$ tends to infinity with $n$.


This is an interesting problem to solve, since the extreme cases (e.g. $(1,\ldots, 1)$ or $(1, 0, \ldots, 0)$) do not provide counterexamples. The coordinates need to be different but not too far apart.


