jmc

Because $3^6=729<2007<2187=3^7$, it is convenient to begin by counting the number of base-3 palindromes with at most 7 digits. There are two palindromes of length 1, namely 1 and 2. There are also two palindromes of length 2, namely 11 and 22. For $n\ge 1$, each palindrome of length $2n+1$ is obtained by inserting one of the digits $0$, $1$, or $2$ immediately after the $n\text{th}$ digit in a palindrome of length $2n$. Each palindrome of length $2n+2$ is obtained by similarly inserting one of the strings $00$, $11$, or $22$. Therefore there are 6 palindromes of each of the lengths 3 and 4, 18 of each of the lengths 5 and 6, and 54 of length 7. Because the base-3 representation of 2007 is 2202100, that integer is less than each of the palindromes 2210122, 2211122, 2212122, 2220222, 2221222, and 2222222. Thus the required total is $2+2+6+6+18+18+54-6=\boxed{100}$.