26. The overline in this problem will exclusively denote binary representation. We will show by induction that if $n = \overline{c_k c_{k-1} \ldots c_0} = \sum_{i=0}^{k} c_i 2^i$ is the binary representation of $n \left( c_i \in \{0,1\} \right)$, then $f(n) = \overline{c_0 c_1 \ldots c_k} = \sum_{i=0}^{k} c_i 2^{k-i}$ is the number whose binary representation is the palindrome of the binary representation of $n$. This evidently holds for $n \in \{1,2,3\}$. Let us assume that the claim holds for all numbers up to $n-1$ and show it holds for $n = \overline{c_k c_{k-1} \ldots c_0}$. We observe three cases: (i) $c_0 = 0 \Rightarrow n = 2m \Rightarrow f(n) = f(m) = \overline{0 c_1 \ldots c_k} = \overline{c_0 c_1 \ldots c_k}$. (ii) $c_0 = 1, c_1 = 0 \Rightarrow n = 4m + 1 \Rightarrow f(n) = 2f(2m + 1) - f(m) = 2 \cdot \overline{1 c_2 \ldots c_k} - \overline{c_2 \ldots c_k} = 2^k + 2 \cdot \overline{c_2 \ldots c_k} - \overline{c_2 \ldots c_k} = \overline{10 c_2 \ldots c_k} = \overline{c_0 c_1 \ldots c_k}$. (iii) $c_0 = 1, c_1 = 1 \Rightarrow n = 4m + 3 \Rightarrow f(n) = 3f(2m + 1) - 2f(m) = 3 \cdot \overline{1 c_2 \ldots c_k} - 2 \cdot \overline{c_2 \ldots c_k} = 2^k + 2^{k-1} + 3 \cdot \overline{c_2 \ldots c_k} - 2 \cdot \overline{c_2 \ldots c_k} = \overline{11 c_2 \ldots c_k} = \overline{c_0 c_1 \ldots c_k}$. We thus have to find the number of palindromes in binary representation smaller than $1998 = \overline{11111000100}$. We note that for all $m \in \mathbb{N}$ the numbers of $2m$- and $(2m-1)$-digit binary palindromes are both equal to $2^{m-1}$. We also note that $\overline{11111011111}$ and $\overline{11111111111}$ are the only 11-digit palindromes larger than 1998. Hence we count all palindromes of up to 11 digits and exclude the largest two. The number of $n \leq 1998$ such that $f(n) = n$ is thus equal to $1 + 1 + 2 + 2 + 4 + 4 + 8 + 8 + 16 + 16 + 32 - 2 = 92$.