Answer: 279

Solution:

If $n$ has a single digit $d,$ then the numeral $d$ has value $d$ in every base, so $f(n) = n:$ all $9$ one-digit numbers work. If $n$ has digits $d_{k-1} \ldots d_1 d_0$ with $k \ge 2,$ then $b \le 10$ always, and if $b \lt 10$ then $f(n) = \sum d_i b^i \lt \sum d_i 10^i = n$ because the leading digit satisfies $d_{k-1} b^{k-1} \lt d_{k-1} 10^{k-1}.$ So a multi-digit $n$ satisfies $f(n) = n$ exactly when $b = 10,$ that is, when some digit of $n$ equals $9.$

Two-digit numbers containing a $9:$ the numbers $90$ through $99$ plus $19, 29, \ldots, 89,$ for $10 + 8 = 18.$ Three-digit numbers containing a $9:$ $900 - 8 \cdot 9 \cdot 9 = 252,$ subtracting the numbers with no $9$ (leading digit $1$–$8,$ others $0$–$8$).

The total is $9 + 18 + 252 = 279.$