Solution 1. For every positive integer $n$, let $M_{n}=\max \left(a_{1}, a_{2}, \ldots, a_{n}\right)$. We first prove that
$$\frac{3^{a_{1}}+3^{a_{2}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}}\right)^{2}} \leqslant\left(\frac{3}{4}\right)^{M_{n}}$$
For $i=1,2, \ldots, n$, from $\left(\frac{3}{2}\right)^{a_{i}} \leqslant\left(\frac{3}{2}\right)^{M_{n}}$ we can obtain $3^{a_{i}} \leqslant\left(\frac{3}{4}\right)^{M_{n}} \cdot 2^{M_{n}} \cdot 2^{a_{i}}$. By summing up over all $i$,
$$\sum_{i=1}^{n} 3^{a_{i}} \leqslant\left(\frac{3}{4}\right)^{M_{n}} \cdot 2^{M_{n}} \cdot \sum_{i=1}^{n} 2^{a_{i}} \leqslant\left(\frac{3}{4}\right)^{M_{n}} \cdot\left(\sum_{i=1}^{n} 2^{a_{i}}\right)^{2}$$
which is equivalent to the previous inequality.
Now let $\mu=\log _{4 / 3} \frac{1}{\varepsilon}$, so that $\mu$ is the positive real number with $\left(\frac{3}{4}\right)^{\mu}=\varepsilon$. If there is an index $n$ such that $a_{n}>\mu$, then $M_{n} \geqslant a_{n}>\mu$, and hence
$$\frac{3^{a_{1}}+3^{a_{2}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}}\right)^{2}} \leqslant\left(\frac{3}{4}\right)^{M_{n}}<\left(\frac{3}{4}\right)^{\mu}=\varepsilon .$$
Otherwise we have $0<a_{i} \leqslant \mu$ for all positive integers $i$, so
$$\frac{3^{a_{1}}+3^{a_{2}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}}\right)^{2}} \leqslant \frac{n \cdot 3^{\mu}}{(n \cdot 1)^{2}}=\frac{3^{\mu}}{n}$$
If $n>\left\lfloor\frac{3^{\mu}}{\varepsilon}\right\rfloor$, this is less than $\varepsilon$.

Solution 2. We will combine two upper bounds.
First, start with the trivial estimate
$$\frac{3^{a_{1}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+\cdots+2^{a_{n}}\right)^{2}} \leqslant \frac{3^{a_{1}}+\cdots+3^{a_{n}}}{4^{a_{1}}+\cdots+4^{a_{n}}}$$
By applying Jensen's inequality to the convex function $x^{\log _{3} 4}$ we get
$$\frac{4^{a_{1}}+\cdots+4^{a_{n}}}{n}=\frac{\left(3^{a_{1}}\right)^{\log _{3} 4}+\cdots+\left(3^{a_{n}}\right)^{\log _{3} 4}}{n} \geqslant\left(\frac{3^{a_{1}}+\cdots+3^{a_{n}}}{n}\right)^{\log _{3} 4}$$
so
$$\frac{3^{a_{1}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+\cdots+2^{a_{n}}\right)^{2}} \leqslant \frac{3^{a_{1}}+\cdots+3^{a_{n}}}{4^{a_{1}}+\cdots+4^{a_{n}}} \leqslant\left(\frac{n}{3^{a_{1}}+\cdots+3^{a_{n}}}\right)^{\log _{3} 4-1}$$
Hence, the original inequality holds true whenever
$$3^{a_{1}}+\cdots+3^{a_{n}}>\left(\frac{1}{\varepsilon}\right)^{\frac{1}{\log _{3} 4-1}} \cdot n$$
Second, trivially
$$\frac{3^{a_{1}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+\cdots+2^{a_{n}}\right)^{2}} \leqslant \frac{3^{a_{1}}+\cdots+3^{a_{n}}}{n^{2}}$$
so the original inequality is satisfied if
$$3^{a_{1}}+\cdots+3^{a_{n}}<\varepsilon \cdot n^{2}$$
If $n>\left(\frac{1}{\varepsilon}\right)^{1+\frac{1}{\log _{3} 4-1}}$ then $\left(\frac{1}{\varepsilon}\right)^{\frac{1}{\log _{3} 4-1}} \cdot n<\varepsilon \cdot n^{2}$, and therefore at least one of the two conditions is satisfied.

Solution 3. Define $C=\log _{4 / 3} \frac{2}{\varepsilon}$, so that if $a_{i}>C$ then $3^{a_{i}}<\frac{\varepsilon}{2} \cdot 4^{a_{i}}$. We divide the sequence into "small" and "large" terms by how they compare to $C$ : let
$$\mathcal{S}_{n}=\left\{i \leqslant n \mid a_{i} \leqslant C\right\} \quad \text{ and } \quad \mathcal{L}_{n}=\left\{i \leqslant n \mid a_{i}>C\right\}$$
Then the original inequality is equivalent to
$$\frac{\sum_{i \in \mathcal{S}_{n}} 3^{a_{i}}}{\left(\sum_{i \in \mathcal{S}_{n}} 2^{a_{i}}+\sum_{i \in \mathcal{L}_{n}} 2^{a_{i}}\right)^{2}}+\frac{\sum_{i \in \mathcal{L}_{n}} 3^{a_{i}}}{\left(\sum_{i \in \mathcal{S}_{n}} 2^{a_{i}}+\sum_{i \in \mathcal{L}_{n}} 2^{a_{i}}\right)^{2}}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}$$
If $\mathcal{L}_{n}$ is nonempty, we have
$$\frac{\sum_{i \in \mathcal{L}_{n}} 3^{a_{i}}}{\left(\sum_{i \in \mathcal{S}_{n}} 2^{a_{i}}+\sum_{i \in \mathcal{L}_{n}} 2^{a_{i}}\right)^{2}}<\frac{\varepsilon}{2} \cdot \frac{\sum_{i \in \mathcal{L}_{n}} 4^{a_{i}}}{\left(\sum_{i \in \mathcal{L}_{n}} 2^{a_{i}}\right)^{2}} \leqslant \frac{\varepsilon}{2}$$
and this also trivially holds when $\mathcal{L}_{n}$ is empty (in which case the LHS is zero).
Now suppose that $n \geqslant \frac{2}{\varepsilon}\left(\frac{3}{2}\right)^{C}$. Note that $3^{a_{i}} \leqslant\left(\frac{3}{2}\right)^{C} 2^{a_{i}}$ for $i \in \mathcal{S}_{n}$, so we have
$$\frac{\sum_{i \in \mathcal{S}_{n}} 3^{a_{i}}}{\left(\sum_{i \in \mathcal{S}_{n}} 2^{a_{i}}+\sum_{i \in \mathcal{L}_{n}} 2^{a_{i}}\right)^{2}} \leqslant \frac{\left(\frac{3}{2}\right)^{C} \sum_{i \in \mathcal{S}_{n}} 2^{a_{i}}}{\left(\sum_{i \in \mathcal{S}_{n}} 2^{a_{i}}+\sum_{i \in \mathcal{L}_{n}} 2^{a_{i}}\right)^{2}} \leqslant \frac{\left(\frac{3}{2}\right)^{C}}{\sum_{i \in \mathcal{S}_{n}} 2^{a_{i}}+\sum_{i \in \mathcal{L}_{n}} 2^{a_{i}}}<\frac{\left(\frac{3}{2}\right)^{C}}{n} \leqslant \frac{\varepsilon}{2},$$
so we have the original inequality.

Solution 4. For every index $i=1,2, \ldots, n$, apply the weighted AM-GM inequality to numbers $2^{a_{i}}$ and $(n-1)$ with weights $\log _{2} \frac{3}{2} \approx 0.585$ and $\log _{2} \frac{4}{3} \approx 0.415$ as
$$\begin{gathered} 2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}} \geqslant 2^{a_{i}}+(n-1) \\ >\log _{2} \frac{3}{2} \cdot 2^{a_{i}}+\log _{2} \frac{4}{3} \cdot(n-1) \geqslant\left(2^{a_{i}}\right)^{\log _{2} \frac{3}{2}} \cdot(n-1)^{\log _{2} \frac{4}{3}} \\ =\left(\frac{3}{2}\right)^{a_{i}} \cdot(n-1)^{\log _{2} \frac{4}{3}}>\left(\frac{3}{2}\right)^{a_{i}} \cdot(n-1)^{2 / 5} . \end{gathered}$$
By summing up for $i=1,2, \ldots, n$,
$$\left(2^{a_{1}}+\cdots+2^{a_{n}}\right)^{2}=\sum_{i=1}^{n} 2^{a_{i}}\left(2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}}\right)>(n-1)^{2 / 5} \sum_{i=1}^{n} 3^{a_{i}}$$
so
$$\frac{3^{a_{1}}+3^{a_{2}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}}\right)^{2}}<\frac{1}{(n-1)^{2 / 5}}$$
If $n \geqslant\left(\frac{1}{\varepsilon}\right)^{5 / 2}+1$ then $\frac{1}{(n-1)^{2 / 5}}<\varepsilon$.