The condition is equivalent to requiring that the binary representation of $t+i$ has $1$'s in all the places the binary representation that $2a_i$ does. We start by showing that they must differ by exactly a single $1$: Claim: Let $s_2(m)$ denote the number of $1$'s in the binary representation of $m$. Then we have the identity \[ s_2(t) + s_2(t+1) + \dots + s_2(2t) = s_2(0) + s_2(1) + \dots + s_2(t-1) + t. \]Proof. Follows directly by induction on $t$. $\blacksquare$ Since we need $s(t+i) \ge s(2a_i) + 1$ for each $i$, summing across $i$ implies that all the inequalities must be tight, i.e.\ we need $s(t+i) = s(2a_i) + 1$ for all $i$. In other words, we have derived that: Claim: A permutation $(a_0, \dots, a_{t-1})$ works if and only if, for each $i$, the number $2a_i$ is obtained by replacing a single $1$ in the binary representation of $t+i$ by a $0$. Henceforth we denote by $2^k$ the unique power of $2$ in the interval $[t, 2t-1]$. There are two classes of numbers we can pin down directly. If $t+i$ is odd then $2a_i = t+i-1$ is forced. (That's because $t+i$ ends with $1$, but $2a_i$ doesn't, so the last bit must be toggled.) If $t+i$ is even and $t+i \ge 2^k$ then $2a_i = t+i-2^k$ is forced. This follows by induction on $i \ge 2^k-t$, with the base case being clear. At each step, only toggling the largest $2^k$ bit results in a possible choice of $2a_i$ that hasn't been forced in a previous step. For example, when $t=19$, the table looks like the following. The first case is colored in black and the second case is colored in blue; the unallocated entries are colored red. \begin{align*} &\begin{array}{r|ccccc ccccc ccc} t+i & 19 & {\color{red}20} & 21 & {\color{red}22} & 23 & {\color{red}24} & 25 & {\color{red}26} & 27 & {\color{red}28} & 29 & {\color{red}30} & 31 \\ 2a_i & 18 & & 20 && 22 && 24 && 26 && 28 && 30 \\ \end{array} \\ &\begin{array}{r|ccccc ccccc ccccc} t+i & {\color{blue}32} & 33 & {\color{blue}34} & 35 & {\color{blue}36} & 37 \\ 2a_i & {\color{blue}0} & 32 & {\color{blue}2} & 34 & {\color{blue}4} & 36 \end{array} \end{align*}Hence, it remains to show there is a unique way to pair up the remaining even values of $t+i$ colored in red above (which range from $2\left\lceil t/2 \right\rceil$ to $2^k-2$ inclusive) with the remaining values of $2a_i$ (which range from to $2t-2^k$ to $2t - 2 \left\lceil t/2 \right\rceil - 2$ inclusive). So from now on focus only on those red columns. It turns out that this can be done via induction on $t$ using the following idea: replace every number $x$ with $2^k-2-x$ in both rows. Then, this actually corresponds to valid table for the even entries when $t' = 2^k-2-2\left\lceil t/2 \right\rceil$, but upside-down! For example, when $t = 19$ we get $t' = 12$; so the permutation of $b_i$'s corresponding to $t' = 12$ gives the construction for $t = 19$. (And conversely, uniqueness of the $a_i$'s follows from uniqueness of $b_i$'s.) \[ \begin{array}{r|cccccc} 2b_i & 10 & 8 & 6 & 4 & 2 & 0\\ t'+b_i & 12 & 14 & 22 & 20 & 18 & 16 \end{array} \xrightarrow{30-x} \begin{array}{r|cccccc} t+a_i & {\color{red}20} & {\color{red}22} & {\color{red}24} & {\color{red}26} & {\color{red}28} & {\color{red}30} \\ 2a_i & 18 & 16 & 8 & 10 & 12 & 14 \end{array} \]

For two integers $a, b$, let $a \succ b$ if every binary digit of $a$ is pairwise $\ge$ then every binary digit of $b$, and $a \ne b$. If $a \succ b$, define the difference of $a$ and $b$ as the number of digits in $a$ that are $1$ that are $0$ in $b$. Then our goal is to show that there exists a unique pairing of $A = \{t, t+1, \dots, 2t-1\}$ to $B = \{0, 2, \dots, 2t-2\}$ such that the former in each pair $\succ$s the later. We local spam to get a matching which has difference $1$ in each amount. Claim: If $2k+1 \in A, 2k \in B$, then $(2k+1, 2k)$ can be a pair. The idea is that if $(2k+1, 2l), (c, 2k)$ is in a configuration, then $(2k+1, 2k), (c, 2l)$ also works. Applying this, it becomes equivalent to do the following after the stripping ending digits: If $t = 2k$, it then remains to match $A = \{0, 1, 2, \dots, k-1\}$ and $B = \{k, k+1, \dots, 2k-1\}$. If $2^a < 2k-1$ for maximal $a$, then we can pairwise pair off $2^a, \dots, 2k-1$ with $0, \dots, 2k-1 - 2^a$ with difference $1$ to reduce to matching $A = \{2^a-2u, \dots, 2^a-u-1\}$ to $B = \{2^a-u \dots, 2^a-1\}$ for some $u$. However, $a \succ b \iff 2^c - b \succ 2^c - a$ for $2^c \ge a, b$, so this just also reduces to matching $1, \dots, u$ to $u+1, \dots, 2u$, where we inductively get a difference $1$ matching. If $t = 2k-1$, it then remains to match $0, 1, 2, \dots, k-2$ with $k, k+1, k+2, \dots, 2k-2$. It likewise becomes equivalent to match up with $A=\{k-1-u, k-2\}$ with $B = \{2^a-u, 2^a-1\}$ where $u$ is taken such that $k-2 + 2 = 2^a-u$. Then we can once again inductively reduce downwards. Anyways, this shows the existence of a permutation $P$ with difference $1$. However, this implies that no other permutation can exist, because if $a \succ d$ with difference $1$, we can repeatedly replace $(a, b), (c, d)$ with $(a,c), (b,d)$ while remaining valid until we end up at $P$. This gives us a series of permutations $S \mapsto S_1 \mapsto \dots \mapsto S_{k-1} \mapsto P$. Except $P$ has difference $1$, so $S_{k-1}$ can't actually exist, contradiction.

This looks different from the official solution, hopefully this is not a fakesolve. Say a set $A$ covers a set $B$ if there is a unique bijection from $A$ to $B$ where if $a,b$ are paired, then $\binom{a}{b}$ is odd. We want to show that $t+1,t+2,\dots , 2t+1$ covers $0,2,\dots ,2t$. By Lucas' theorem, $\binom{a}{b}$ is odd iff the binary ones of $b$ are a subset of the binary ones of $a$. We now stop doing NT. Claim: In a covering of $0,2,\dots, 2t$ by $t+1,t+2,\dots, 2t+1$, each $t+i$ has exactly one more binary one than $2a_i$. Proof: It suffices to show $t+1,t+2,\dots, 2t+1$ has $t+1$ more binary ones than $0,2,\dots, 2t$. We induct with base case $t=0$, which is easy. If the claim is true for $t-1$, then it suffices to show that the number of binary ones in $2t+1$ and $2t$ is one more than the number of binary ones in $2t$ and $t$. This is true because $2t+1$ in binary is a $1$ added to $t$ in binary. Thus the odd numbers $t+i$ in $t+1,t+2,\dots, 2t+1$ must have the ending $1$ removed. If $t$ is even then $t+1,t+3,\dots ,2t+1$ cover $t,t+2,\dots , 2t$ respectively, and then we want $t+2,t+4,\dots ,2t$ to cover $0,2,\dots t-2$, which by Lucas means we need $\frac{t+2}{2},\dots ,t$ to cover $0,\dots ,\frac{t-2}{2}$. If $t$ is odd then let $t+2,t+4,\dots ,2t+1$ cover $t+1,t+3,\dots ,2t$ respectively, and then we want $t+1,t+3,\dots 2t$ to cover $0,2,\dots ,t-1$, which by Lucas means we need $\frac{t+1}{2},\dots ,t$ to cover $0,\dots ,\frac{t-1}{2}$. Thus the problem is equivalent to the following claim. Claim: $k+1,k+2,\dots ,2k+1$ and $k+2,k+3,\dots ,2k+2$ cover $0,1,\dots ,k$ for $k\ge 0$. Proof: We already know that in a covering of $0,1,\dots, k$ by $k+1,k+2,\dots, 2k+1$ or $k+2,k+3,\dots, 2k+2$, one binary one from every large number must be removed. Strong induct on $k$ with base case $k=0$, which is easy. For the inductive step, we have $4$ things to prove which are basically the same idea, with an extra step in the last case. $k+2,\dots ,2k+2$ covers $0,\dots ,k$ for $k$ odd.

$k+2,\dots ,2k+2$ covers $0,\dots ,k$ for $k$ even.

$k+1,\dots ,2k+1$ covers $0,\dots ,k$ for $k$ odd.

$k+1,\dots ,2k+1$ covers $0,\dots ,k$ for $k$ even.

Nice combi-NT problem! Here's the breakdown of my solution.

v_Enhance wrote: The condition is equivalent to requiring that the binary representation of $t+i$ has $1$'s in all the places the binary representation that $2a_i$ does. We start by showing that they must differ by exactly a single $1$: Claim: Let $s_2(m)$ denote the number of $1$'s in the binary representation of $m$. Then we have the identity \[ s_2(t) + s_2(t+1) + \dots + s_2(2t) = s_2(0) + s_2(1) + \dots + s_2(t-1) + t. \]Proof. Follows directly by induction on $t$. $\blacksquare$ Since we need $s(t+i) \ge s(2a_i) + 1$ for each $i$, summing across $i$ implies that all the inequalities must be tight, i.e.\ we need $s(t+i) = s(2a_i) + 1$ for all $i$. In other words, we have derived that: Claim: A permutation $(a_0, \dots, a_{t-1})$ works if and only if, for each $i$, the number $2a_i$ is obtained by replacing a single $1$ in the binary representation of $t+i$ by a $0$. Henceforth we denote by $2^k$ the unique power of $2$ in the interval $[t, 2t-1]$. There are two classes of numbers we can pin down directly. If $t+i$ is odd then $2a_i = t+i-1$ is forced. (That's because $t+i$ ends with $1$, but $2a_i$ doesn't, so the last bit must be toggled.) If $t+i$ is even and $t+i \ge 2^k$ then $2a_i = t+i-2^k$ is forced. This follows by induction on $i \ge 2^k-t$, with the base case being clear. At each step, only toggling the largest $2^k$ bit results in a possible choice of $2a_i$ that hasn't been forced in a previous step. For example, when $t=19$, the table looks like the following. The first case is colored in black and the second case is colored in blue; the unallocated entries are colored red. \begin{align*} &\begin{array}{r|ccccc ccccc ccc} t+i & 19 & {\color{red}20} & 21 & {\color{red}22} & 23 & {\color{red}24} & 25 & {\color{red}26} & 27 & {\color{red}28} & 29 & {\color{red}30} & 31 \\ 2a_i & 18 & & 20 && 22 && 24 && 26 && 28 && 30 \\ \end{array} \\ &\begin{array}{r|ccccc ccccc ccccc} t+i & {\color{blue}32} & 33 & {\color{blue}34} & 35 & {\color{blue}36} & 37 \\ 2a_i & {\color{blue}0} & 32 & {\color{blue}2} & 34 & {\color{blue}4} & 36 \end{array} \end{align*}Hence, it remains to show there is a unique way to pair up the remaining even values of $t+i$ colored in red above (which range from $2\left\lceil t/2 \right\rceil$ to $2^k-2$ inclusive) with the remaining values of $2a_i$ (which range from to $2t-2^k$ to $2t - 2 \left\lceil t/2 \right\rceil - 2$ inclusive). So from now on focus only on those red columns. It turns out that this can be done via induction on $t$ using the following idea: replace every number $x$ with $2^k-2-x$ in both rows. Then, this actually corresponds to valid table for the even entries when $t' = 2^k-2-2\left\lceil t/2 \right\rceil$, but upside-down! For example, when $t = 19$ we get $t' = 12$; so the permutation of $b_i$'s corresponding to $t' = 12$ gives the construction for $t = 19$. (And conversely, uniqueness of the $a_i$'s follows from uniqueness of $b_i$'s.) \[ \begin{array}{r|cccccc} 2b_i & 10 & 8 & 6 & 4 & 2 & 0\\ t'+b_i & 12 & 14 & 22 & 20 & 18 & 16 \end{array} \xrightarrow{30-x} \begin{array}{r|cccccc} t+a_i & {\color{red}20} & {\color{red}22} & {\color{red}24} & {\color{red}26} & {\color{red}28} & {\color{red}30} \\ 2a_i & 18 & 16 & 8 & 10 & 12 & 14 \end{array} \] The 1st claim is wrong a little bit