Clearly $m$ must be even, or the sum would be negative. Furthermore, if $a_{m} \leq 2013$, the sum cannot exceed $3^{2013}$ since $j=3^{a_{m}}+\sum_{k=0}^{m-1}\left((-1)^{k} \cdot 3^{a_{k}}\right) \leq 3^{a_{m}}$. Likewise, if $a_{m}>2013$, then the sum necessarily exceeds $3^{2013}$, which is not hard to see by applying the Triangle Inequality and summing a geometric series. Hence, the elements of $\left\{a_{k}\right\}$ can be any subset of $\{0,1, \ldots, 2013\}$ with an odd number of elements. Since the number of even-sized subsets is equal to the number of odd-sized elements, there are $\frac{2^{2014}}{2}=2^{2013}$ such subsets. Now, it suffices to show that given such an $\left\{a_{k}\right\}$, the value of $j$ can only be obtained in this way. Suppose for the sake of contradiction that there exist two such sequences $\left\{a_{k}\right\}_{0 \leq k \leq m_{a}}$ and $\left\{b_{k}\right\}_{0 \leq k \leq m_{b}}$ which produce the same value of $j$ for $j$ positive or negative, where we choose $\left\{a_{k}\right\},\left\{b_{k}\right\}$ such that $\min \left(m_{a}, m_{b}\right)$ is as small as possible. Then, we note that since $3^{a_{0}}+3^{a_{1}}+\ldots+3^{\left(a_{m_{a}}-1\right)} \leq 3^{0}+3^{1}+\ldots+3^{\left(a_{m_{a}-1}\right)}<2\left(3^{\left(a_{m_{a}}-1\right)}\right)$, we have that $\sum_{k=0}^{m_{a}}\left((-1)^{k} \cdot 3^{a_{k}}\right)>3^{\left(a_{m_{a}}-1\right)}$. Similarly, we get that $3^{\left(a_{m_{b}}-1\right)} \geq \sum_{k=0}^{m_{b}}\left((-1)^{k} \cdot 3^{a_{k}}\right)>3^{\left(m_{b}-1\right)}$; for the two to be equal, we must have $m_{a}=m_{b}$. However, this means that the sequences obtained by removing $a_{m_{a}}$ and $a_{m_{b}}$ from $\left\{a_{k}\right\}\left\{b_{k}\right\}$ have smaller maximum value but still produce the same alternating sum, contradicting our original assumption.