Let $t>0$ and $\delta\in\big(0,\frac12\big)$ be fixed. For any $k\in\mathbb{N}$ let $I\_k,J\_k\in\mathbb{N}$ be finite subsets of natural numbers with cardinalities denoted as $|I\_k|,|J\_k|$, respectively. Now define numbers


$\hspace{70pt}C\_k:=\max\Big\{\max\limits\_{i\in I\_k}i^{2t}\sum\limits\_{j\in J\_k}j^{2t},\,\,\max\limits\_{j\in J\_k}j^{2t}\sum\limits\_{i\in I\_k}i^{2t}\Big\}$


and


$\hspace{70pt}D\_k:=\max\Big\{|I\_k|\max\limits\_{j\in J\_k}j^{4t+2\delta},\,\,|J\_k|\max\limits\_{i\in I\_k}i^{4t+2\delta}\Big\}$.


Aim: choose the subsets $|I\_k,|J\_k|$ so that


$\hspace{100pt}R\_k:=\frac{C\_k}{D\_k}\to\infty\,\,\,\,$as$\,\,\,\,k\to\infty$.



Choosing finite subsets of natural numbers
As @Krzysztof noticed, there was a mistake in the original post, which was that I was looking at $2t+2\delta$ and perceived it (for some unknown reason) as $2t(1+\delta)$. So here is the correctedversion:


Let $I,J$ be the interval lengths and $i,j$ be the top numbers in the intervals.
Then the sum of integers in $I$ to the power $s>0$ is $i^sI$ up to a constant factor.
What you want is 
$$
i^{2t}j^{2t}J\gg \max(Ij^{4t+2\delta},Ji^{4t+2\delta})\,
$$
i.e.,
$i^{2t}J\ge Ij^{2t+2\delta}$ and $j^{2t}\gg i^{2t+2\delta}$\,.
Clearly, we should take $I=1$ to make the task easier. The second inequality gives $i^{2t}\ll j^{2t\frac t{t+\delta}}$, so, plugging it into the first inequality, we see that we want
$$
J\gg j^{2t[\frac{t+\delta}t-\frac t{t+\delta}]}\,.
$$
This is compatible with the trivial restriction $j\ge J$ for large $j$ if and only if the power on the RHS is less than $1$.


