Yes. Is is easy to prove it with $\epsilon-N_\epsilon$, but here is a simpler proof:
Assume by contradiction that some rearangement doesn't diverge to $\infty$. Then the re-arangement is convergent (since positive) and hence absolutely convergent.
But then, any re-arangement of the re-arangement is convergent to the same real number... The original series is a re-arangement of any of his re-arangements....
P.S. Here is the $\epsilon-N_\epsilon$ proof.
Let $\sigma$ be any permutation of the indices. We will denote by $s_n$ respectively $t_n$ the partial sums of the original series and the permutation.
Let $\epsilon >0$ be arbitrary.
Since $\lim_n s_n =\infty$ then, there exists some $N_\epsilon$ so that for all $n > N_\epsilon$ we have
$s_n > \epsilon \,.$
Pick now some $M_\epsilon$ such that
$\{1,2,3,.., N_\epsilon\} \subset \sigma( \{ 1,2,...., M_\epsilon \}) $
Then
$t_{M_\epsilon}=x_{\sigma(1)}+....+x_{\sigma(M_\epsilon)} \geq x_1+...+x_{N_\epsilon} =s_{N_\epsilon}$
Thus, for all $n> M_\epsilon$ we have
$T_n \geq t_{M_\epsilon} \geq s_{N_\epsilon} > \epsilon \,.$
Since $\epsilon >0$ is arbitrary we are done.