Solution:

Sketch of solution: Initially the number $a$ is written on the board. The transition from $x$ to $2x+1$ or $\frac{x}{x+2}$ shall be called a transformation. All numbers on the board are positive.

1st variant:
From the number $a$, after $k$ transformations one always obtains numbers of the form $\frac{ma+m-1}{(2^{k}-m)a+2^{k}-m+1}$ with an integer $0 < m \leq 2^{k}$ (proof by complete induction on $k$). So if 2008 appears on the board at some point, there exist $k \geq 0$ and $0 < m \leq 2^{k}$ with $\frac{ma+m-1}{(2^{k}-m)a+2^{k}-m+1} = 2008$, i.e. $2009 = (a+1)(2009m - 2008 \cdot 2^{k})$. Since the second factor is coprime to 2009 (as 2008 and $2^{k}$ are coprime to 2009), it follows that $a+1 = 2009$.

2nd variant:
On a sheet of paper, write for each number $t$ on the board the number $f(t) = \frac{2009}{t+1}$. At some point 1 appears on the sheet. A transformation of a number on the board maps a number $z$ on the sheet to $z/2$ or $\frac{2009+z}{2}$. So if the initial number $f(a)$ on the sheet is not an integer or not coprime to 2009, then integers, or numbers coprime to 2009, can never arise from it. Hence $f(a)$ is an integer and coprime to 2009. Since $f(a) = \frac{2009}{a+1}$ also divides 2009, it follows that $f(a) = 1$ and $a = 2008$.

3rd variant:
Every transformation maps the reduced rational number $p/q$ to the reduced number $p'/q'$, where $p, q, p', q'$ are positive integers and $p'+q' \in \{p+q, 2(p+q)\}$. Initially one has the reduced representation $\frac{a}{1}$, at the end $\frac{2008}{1}$, so there exists $k \geq 0$ with $2^{k}(a+1) = 2008+1$. Since 2009 is odd, it follows that $k=0$ and $a+1 = 2009$.

4th variant:
Form the sequence of numbers $a_{0} = a, a_{1}, \ldots, a_{k} = 2008$ from numbers on the board, where $a_{j+1}$ for $0 \leq j < k$ arises from $a_{j}$ by one of the two transformations. From $a_{k}$ one can uniquely reconstruct $a_{k-1}, \ldots, a_{0}$, since for all $x > 0$ we have: $2x+1 > 1$, $0 < \frac{x}{x+2} < 1$. By induction on $n$ for $n = 0, \ldots, k$ one shows that $a_{k-n} = \frac{2009-b_{n}}{b_{n}}$ with $0 < b_{n} < 2009$ and $b_{n} \equiv 2^{n} \bmod 2009$. Thus $a_{k-n}$ is an integer (with value 2008) exactly when $2^{n} \equiv 1 \bmod 2009$, i.e. when $420 \mid n$.