The answer is 4.

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Relevant wiki: Arithmetic ProgressionsLet the number of terms be $n$ and the common difference be $d$

We know that $T_n = 2016$

$1729 + (n-1)(d) = 2016\\ (n-1)(d) = 287\\ (n-1)(d) = 1 \times 7 \times 41$

Since both $(n-1)$ and $d$ are positive integers, we have these possible combinations:

$n-1=1,\quad d= 7 \times 41\\ n-1=7,\quad d= 41\\ n-1=41,\quad d= 7\\ n-1=7 \times 41,\quad d=1$

Therefore, there are $\boxed{4}$ possible arithmetic progressions