We want $X$ to be a binomially distributed random variable $X\sim \text{Bin}(n,p)$, where $p=2/27$. (Thus $\mathbb{E}(X)=\frac{2n}{27}$ and $\mathbb{V}(X)=\frac{50n}{27^2}$.) What is the smallest number of bernoulli repetitions $n$ that would allow us to use Chebychev's inequality (CI) to conclude that $P(X>0)\geq 0.25$?
Chebychev's inequality: $$P(|X-\mathbb{E}(X)|\geq \epsilon)\leq \frac{\mathbb{V}(X)}{\epsilon^2}, \text{ where } \epsilon >0 \,.$$
My approach:
We can equivalently look at $P(X\leq 0)<0.75$ and realize $$ P(X\leq 0)=P(\mathbb{E}(X)-X\geq \mathbb{E}(X))\leq P(|\mathbb{E}(X)-X|\geq \mathbb{E}(X)). $$ The upper bound provided by CI yields the condition $$ \frac{50n}{27^2\epsilon^2}<0.75\implies n<10.935\epsilon^2. $$ In order to apply the CI at $P(X\leq 0)$ we have to consider $0<\epsilon\leq\mathbb{E}(X)$ because then
$$ P(X\leq 0)\leq P(|\mathbb{E}(X)-X|\geq \mathbb{E}(X))\leq P(|\mathbb{E}(X)-X|\geq \epsilon). $$ So we get a second condition $$ 0<\epsilon\leq \mathbb{E}(X)=\frac{2n}{27}\implies \frac{27\cdot \epsilon}{2}\leq n. $$ In order to find the minimum value of $n$ we have to consider possible $n\in\mathbb{N}$ such that $$ \frac{27\cdot \epsilon}{2}\leq n<10.935\epsilon^2. $$ My conclusion is that $n=17$.
This approach feels a bit messy. Is it correct? Is there a more elegant approach using CI to obtain a lower bound on $n$? Maybe I can avoid the interval at the end of my proof or maybe there is some magic trick when applying CI to get a smaller $n$?
Most of your solution seems to be correct and straightforward. You arrive at the correct answer, but the final step is a bit vague and hopefully I can help to make the reasoning clearer.
Logically speaking, I would like to say that $n$ has to be a fixed number, although unknown, before we can use Chebychev's inequality on the random variable $X$ having the parameter $n$! Let us regard $n$ as fixed for now; we can evaluate afterwards whether we "fixed the right $n$".
By rewriting the left hand side (I recommend drawing some intervals on the real line) Chebychev's inequality can be formulated
$$P(\{X\leq \mathbb{E}(X) - \epsilon\} \cup \{ X \geq \mathbb{E}(X) + \epsilon\} \leq \frac{\mathbb{V}(X)}{\epsilon^2}\,.$$
Since $X$ is nonnegative, $P(X\leq 0)= P(X=0)$, and so we want to be able to use CI to conclude
$$ P(X=0) \leq \frac{3}{4}\,.$$
You rightly note that in order to use CI to get an upper bound on $P(X=0)$ it is necessary to have $\epsilon\leq\mathbb{E}(X)$. While respecting this, we do want to have $\epsilon$ as large as possible to make the right hand side of Chebychev as small as possible. Therefore we choose $\epsilon = \mathbb{E}(X)$.
With $\epsilon = \mathbb{E}(X) = 2n/27$, CI implies
$$P(\{X = 0\} \cup \{ X \geq 2\mathbb{E}(X)\}) \leq \frac{25}{2n}\,.$$
Combining the above equation with
$$P(\{X=0\}) \leq P(\{X = 0\} \cup \{ X \geq 2\mathbb{E}(X)\})$$
we obtain
$$ P(X=0) \leq \frac{25}{2n}\,.$$
In order for the right hand side to be less than 3/4 we need $n\geq 17$ which gives the answer $n=17$.
Note that it is possible to calculate $P(X=0)$ directly by $\binom{n}{0}(1-p)^n$. This gives $$P(X=0)\leq \frac{3}{4} \Longleftrightarrow \left(\frac{25}{27}\right)^n \leq \frac{3}{4}\,,$$
which holds true already for $n\geq 4$.