7. If 2008 numbers $a_{1}, a_{2}, \cdots, a_{2008}$ satisfy
$$
a_{1}=2, a_{n}^{2}-\left(\frac{a_{n-1}}{2008}+\frac{1}{a_{n-1}}\right) a_{n}+\frac{1}{2008}=0,
$$

where, $n=2,3, \cdots, 2008$, then the maximum value that $a_{2008}$ can reach is
7. $\frac{2008^{2006}}{2}$.

From the given information, we have
$$
a_{n}=\frac{1}{2} \frac{a_{n \cdot 1}}{2008} \text { or } a_{n} \stackrel{1}{2}=\frac{1}{a_{n \cdot 1}} \text {, }
$$

This means that $a_{n-1}$ can only be transformed into $a_{n}(n=2,3, \cdots, 2008)$ through either the (1)st type of transformation or the (2)nd type of transformation, starting from $a_{1}$ and undergoing 2007 such transformations to become $a_{2008}$.

Two consecutive (2)nd type transformations cancel each other out, leaving the original number unchanged.

When three consecutive transformations are “(1)st type transformation, (2)nd type transformation, (1)st type transformation,” the two (1)st type transformations cancel each other out, effectively performing only one (2)nd type transformation on the original number. Therefore, 2007 consecutive transformations on 2 are equivalent to performing $m$ consecutive (1)st type or (2)nd type transformations on 2, and the (2)nd type transformation can only occur in the first and last transformations. When 2 undergoes 2007 consecutive transformations: “the first 2006 are (1)st type transformations, and the last one is a (2)nd type transformation,” $a_{2008}$ reaches its maximum value $\frac{2008^{2006}}{2}$.