jmc

We can let the factorization be $$x^2+bx+2008=(x+p)(x+q),$$where $p$ and $q$ are integers. Then $p+q=b$ and $pq=2008.$

The equation $pq=2008$ tells us that either both $p$ and $q$ are positive, or both are negative. Since $p+q=b$ is positive, both $p$ and $q$ are positive.

We want to find the minimum value of $b.$ The number $b=p+q$ is minimized when $p$ and $q$ are as close as possible, under the condition $pq=2008.$ This occurs when $p$ and $q$ are 8 and 251, so the smallest possible value of $b$ is $8+251=\boxed{259}.$