First, you cannot have 3 positive numbers $2^a < 2^b < 2^c$ in $T$ because the difference between $2^b-2^a \neq 2^c-2^b$. Likewise, you cannot have 3 negative numbers. So now you're bound to two positive and two negative numbers at most. We shall show that we cannot have four numbers in $T$, but rather if $T$ has two positive numbers then it can only have one negative number. Let the positive numbers be $2^a < 2^b$, and let $-2^{c}$ be the largest of the negative numbers. $$ 2^b - 2^a = 2^a + 2^c$$$$2^b - 2^c = 2^{a+1}$$This is only possible where $b = c+1$ so that $a = c-1$, so $T = \{-2^{a+1}, 2^a, 2^{a+2}\}, a = 0,1,\dots,2020$ Suppose there is another negative number $-2^d < -2^c$ then we must have: $$2^d - 2^c = 3.2^a$$but because $c=a+1$: $$2^{d-1}-2^a = 3.2^{a-1}$$which means the LHS is divisible by $2^a$ but RHS is not. Contradiction. So $T$ is limited to only 2 positive numbers and 1 negative number, or vice versa. The sets with 2 positive and 1 negative numbers are given by $$T = \{-2^{a+1}, 2^a, 2^{a+2}\}, a = 0,1,\dots,2020$$So there are 2021 sets. The sets with 2 negative and 1 positive numbers are the same but flipped.