The problem is :
$X_1,X_2, \ldots $ are independent rvs with \begin{align*} X_k = \begin{cases} - ke^k \text{ with prob. } e^{-2k} \\ + ke^k \text{ with prob. } e^{-2k} \\ - k \text{ with prob. } \frac{1}{2} - e^{-2k} \\ + k \text{ with prob. } \frac{1}{2} - e^{-2k} \end{cases} \end{align*} Define $S_n := \sum_{k=1}^n X_k$. Can you find constants $a_n>0$ such that $a_nS_n$ converges to a non degenerate distribution ?
Now, after playing around a bit with $\mathbb{E}(X_k)$ and $\mathbb{V}(X_k)$, it seems that the candidates $a_n$ must be of the form $\mathcal{O}(n^{\frac{3}{2}})$ if $a_n S_n$ has the non degenerate distribution. However, I cannot proceed any further.
I would to see how to complete this line of attack, or if this needs to be approached in a different way.
Let $A_k$ denote the event $\{|X_k| > k\}.$ Then $\mathbf{P}(A_k) = 2e^{-2k}$ and the Borel-Cantelli lemma shows that only finitely many of the $A_k$ will occur with probability one. Let $a_n$ denote any sequence of positive real numbers converging to zero. We can write $$ X_k = \Big( k e^k \mathbf{1}_{A_k} + k \mathbf{1}_{A_k^\complement} \Big) (2B_k - 1) = \Big( k (e^k - 1) \mathbf{1}_{A_k} + k \Big) (2B_k - 1), $$ where the $(B_k)$ is a sequence of independent and identically distributed $\mathsf{Ber}(\frac{1}{2})$ random variables.
Since almost surely only finitely many $A_k$ will occur, it easily follows that $$ a_n S_n = o(1) + a_n \sum_{k = 1}^n k (2B_k - 1), $$ where the $o(1)$ term is a random variable that converges to zero a.s. A fortiori, the limits of $a_n S_n$ as the same as those of the sequence $a_n \sum\limits_{k = 1}^n k(2B_k - 1)$ and we can therefore study the latter.
Write $\xi_k = k(2B_k - 1).$ Then, $\mathbf{E}(\xi_k) = 0,$ $|\xi_k| = k$ and $\xi_k^2 = k^2,$ so that $\mathbf{V}(\xi_k) = k^2.$ Let $s_n^2 = \sum\limits_{k = 1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6} = \mathbf{V}(S_n).$ By the Lindeberg-Feller CLT, for the ratio $S_n/s_n$ to converge weakly to a $\mathsf{Norm}(0;1)$ random variable it is necessary and sufficient that, for every $t > 0,$ $$ s_n^{-2} \sum_{k = 1}^n \mathbf{E}(\xi_k^2 \mathbf{1}_{\{|\xi_k| > ts_n\}}) \to 0 $$ as $n \to \infty.$ (This is known as "Linderberg's condition.") Note that, in this case, Lindeberg's condition reduces to $$ s_n^{-2} \sum_{k = 1}^n k^2 \mathbf{1}(k > ts_n). $$ When $t s_n > n$ (which will happen for all $n$ large as the left hand side is dominated by $n^\frac{3}{2}$) there are no index $1 \leq k \leq n$ satisfying the condition in the indicator inside the previous sum and as such the Lindeberg's condition holds. Therefore, $$ \sqrt{\dfrac{6}{n (n+1)(2n+1)}} S_n \to \mathsf{Norm}(0; 1). $$ Finally, to solve your question, you want any sequence $a_n$ such that $a_n s_n \to \sigma > 0,$ this implies $a_n S_n = a_n s_n \dfrac{S_n}{s_n} \to \mathsf{Norm}(0; \sigma^2).$ Observe that such a sequence $a_n$ will have $a_n \sim \sigma/n^\frac{3}{2},$ so the condition imposed above that $a_n \to 0$ actually holds.