jmc

The question implies that we can factor the given quadratic as $$\begin{array}{rl}{x}^2+bx+2008& =(x+r)(x+s)\\ & =x^2+(r+s)x+rs,\end{array}$$ where $r$ and $s$ are integers. Since both $b$ and 2008 are positive, it is clear that $r$ and $s$ must also be positive. By multiplying out the right-hand side as shown, we see that we must have $rs=2008$, which has prime factorization $2008=2\cdot 2\cdot 2\cdot 251$. Recall that we are trying to minimize $b=r+s$. The best we can do is to let $r=251$ and $s=8$, leading to $b=251+8=\boxed{259}$.