Throughout the solution we denote the cells of the table by coordinate pairs; $(i, j)$ refers to the $j$ th cell in the $i$ th row. Consider the directed graph, whose vertices are the cells of the board, and the edges are the arrows $(i, j) \rightarrow(j, i+j)$ for all $0 \leq i, j \leq 2^{n}-1$. From each vertex $(i, j)$, exactly one edge passes $\left(\right.$ to $\left(j, i+j \bmod 2^{n}\right)$ ); conversely, to each cell $(j, k)$ exactly one edge is directed (from the cell $\left.\left(k-j \bmod 2^{n}, j\right)\right)$. Hence, the graph splits into cycles. Now, in any coloring considered, the vertices of each cycle should have the same color by the problem condition. On the other hand, if each cycle has its own color, the obtained coloring obviously satisfies the problem conditions. Thus, the maximal possible number of colors is the same as the number of cycles, and we have to prove that this number is $2^{n}$. Next, consider any cycle $\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right), \ldots$; we will describe it in other terms. Define a sequence $\left(a_{0}, a_{1}, \ldots\right)$ by the relations $a_{0}=i_{1}, a_{1}=j_{1}, a_{n+1}=a_{n}+a_{n-1}$ for all $n \geq 1$ (we say that such a sequence is a Fibonacci-type sequence). Then an obvious induction shows that $i_{k} \equiv a_{k-1}\left(\bmod 2^{n}\right), j_{k} \equiv a_{k}\left(\bmod 2^{n}\right)$. Hence we need to investigate the behavior of Fibonacci-type sequences modulo $2^{n}$. Denote by $F_{0}, F_{1}, \ldots$ the Fibonacci numbers defined by $F_{0}=0, F_{1}=1$, and $F_{n+2}=$ $F_{n+1}+F_{n}$ for $n \geq 0$. We also set $F_{-1}=1$ according to the recurrence relation. For every positive integer $m$, denote by $\nu(m)$ the exponent of 2 in the prime factorization of $m$, i.e. for which $2^{\nu(m)} \mid m$ but $2^{\nu(m)+1} \backslash m$. Lemma 1. For every Fibonacci-type sequence $a_{0}, a_{1}, a_{2}, \ldots$, and every $k \geq 0$, we have $a_{k}=$ $F_{k-1} a_{0}+F_{k} a_{1}$. Proof. Apply induction on $k$. The base cases $k=0,1$ are trivial. For the step, from the induction hypothesis we get
$$a_{k+1}=a_{k}+a_{k-1}=\left(F_{k-1} a_{0}+F_{k} a_{1}\right)+\left(F_{k-2} a_{0}+F_{k-1} a_{1}\right)=F_{k} a_{0}+F_{k+1} a_{1}$$
Lemma 2. For every $m \geq 3$, (a) we have $\nu\left(F_{3 \cdot 2^{m-2}}\right)=m$; (b) $d=3 \cdot 2^{m-2}$ is the least positive index for which $2^{m} \mid F_{d}$; (c) $F_{3 \cdot 2^{m-2}+1} \equiv 1+2^{m-1}\left(\bmod 2^{m}\right)$. Proof. Apply induction on $m$. In the base case $m=3$ we have $\nu\left(F_{3 \cdot 2^{m-2}}\right)=F_{6}=8$, so $\nu\left(F_{3 \cdot 2^{m-2}}\right)=\nu(8)=3$, the preceding Fibonacci-numbers are not divisible by 8 , and indeed $F_{3 \cdot 2^{m-2}+1}=F_{7}=13 \equiv 1+4(\bmod 8)$. Now suppose that $m>3$ and let $k=3 \cdot 2^{m-3}$. By applying Lemma 1 to the Fibonacci-type sequence $F_{k}, F_{k+1}, \ldots$ we get
$$\begin{gathered} F_{2 k}=F_{k-1} F_{k}+F_{k} F_{k+1}=\left(F_{k+1}-F_{k}\right) F_{k}+F_{k+1} F_{k}=2 F_{k+1} F_{k}-F_{k}^{2} \\ F_{2 k+1}=F_{k} \cdot F_{k}+F_{k+1} \cdot F_{k+1}=F_{k}^{2}+F_{k+1}^{2} \end{gathered}$$
By the induction hypothesis, $\nu\left(F_{k}\right)=m-1$, and $F_{k+1}$ is odd. Therefore we get $\nu\left(F_{k}^{2}\right)=$ $2(m-1)>(m-1)+1=\nu\left(2 F_{k} F_{k+1}\right)$, which implies $\nu\left(F_{2 k}\right)=m$, establishing statement (a). Moreover, since $F_{k+1}=1+2^{m-2}+a 2^{m-1}$ for some integer $a$, we get
$$$$ F_{2 k+1}=F_{k}^{2}+F_{k+1}^{2} \equiv 0+\left(