we can easily see that $ 16 | y^2$ so we replace $ y$ by $ 4y$. we have: $ 4z=\frac{49 \times 251}{y^2}-\frac{2007 \times 2009}{x^2}$ since we are trying to find the greatest value of $ z$ we assume that $ z$ is positive and we know that it is an integer which leads to $ y^2 | 251.49.x^2,x^2 | 2007.2009.y^2$ also none of $ 49$ and $ 251$ divides $ y$(note that these numbers are prime) if not,then RHS will be at maximum $ 5,...$(it is attained when $ y=49$ and if $ y=251s$ then RHS will be lesser than $ 1$) and because LHS is divisible by $ 4$ so we must have $ z=1$ and then we'll have $ z=1, \frac{251}{49}-\frac{2007.2009}{x^2}=4 \implies \frac{55}{49}=\frac{2007.2009}{x^2}$ but we don't have $ 55 | 2007.2009.49$ so $ y$ divides $ x$ and then $ yt=x$ so we 'll have $ 251.49.t^2=4zy^2t^2+2007.2009$ from these we can find all possible values of $ t$ since $ t^2 | 2007.2009$ or consequently $ t^2 | 9$(note that $ 2009=49.41$ and $ 2007=9.223$) which implies $ t=1,3$.the rest is easy after finding $ y$ by putting $ t=1,3$.

We claim that the answer is $z = \boxed{789}.$ This is achieved by $x = 21, y = 4.$ We will now show that this is optimal. First of all, it's easy to see by $v_2$ considerations that $4|y.$ If $|y|>4$, then we have $\frac{2z}{2009} = \frac{2008}{41y^2} - \frac{2007}{2x^2} < \frac{2008}{41 \cdot 8^2} < \frac{2 \cdot 789}{2009}$, and so we're done in this case. Hence, assume WLOG that $y = 4$ from now on. We have that: $$\frac{2z}{2009} = \frac{251}{82} - \frac{2007}{2x^2}.$$ Observe that we must have $3|x$ by $v_3$ considerations, and actually that $v_3(x) = 1.$ We also have that $v_7(x) \le 1$. Finally, we know that for any prime $p \neq 3, 7$, $p \nmid x$, as otherwise we'd have that $v_p(RHS) \le -2$, contradiction. Hence, $x \le 21$, and so the RHS is at most $\frac{251}{82} - \frac{2007}{2 \cdot 21^2} = \frac{2 \cdot 789}{2009},$ as desired. $\square$