Here's a way to see both directions roughly at the same time. For any $1\le k\le n$, let $J_k$ be the subregion $\{ x \in J : x_k^{a_k} = \max(x_1^{a_1}, \ldots, x_n^{a_n})\}$. Equivalently, $x \in J_k$ iff $x_i \le x_k^{a_k/a_i}$ for each $1 \le i \le k$.
Notice that for any $x \in J_k$ we have $x_k^{a_k} \le \sum_{i=1}^n x_i^{a_i} \le nx_k^{a_k}.$ Let's define $I_k = \int_{J_k} x_k^{-a_k} dx.$ By the above inequalities, we clearly have $I \ge \frac1n I_k$ since $J_k \subset J$, and also $I \le I_1 + I_2 + \cdots + I_n$, because $J_1 \cup \cdots \cup J_n = J$.
Therefore $I$ converges if and only if each $I_k$ converges. It is now a routine matter to calculate $I_k$ by iterated integration, saving $x_k$ for last. The innermost $n-1$ integrals are just the volume of a box with dimensions $x_k^{a_k/a_i}$ (from the definition of $J_k$). For instance, when $k=1$ we get
$I_1 = \int_0^1 \left(\prod_{i=2}^n x_1^{a_1/a_i} \right) x_1^{-a_1} dx_1 = \int_0^1 x_1^{-a_1 + a_1/a - 1} dx_1,$ where $\frac1a = \frac1{a_1} + \cdots + \frac1{a_n}$ as in @BobTerrell's solution. This one-dimensional integral converges iff $-a_1 + a_1/a > 0$, in other words exactly when $a < 1$.