Biggest G.C.D !!!

First of all, note that:
$999= 3^3×37$
Now, the number of numbers ( sounds funny) is 49 and hence the g.c.d. can be no more than 20 i.e. the biggest integer less than $999/49$
Let their g.c.d be $x$ .
Then the numbers are of the form:
$ax + bx + ..................+ Wx$ $(49 numbers)$
Their total will be: $(a+b+c+.............+W)$ $* x$
Hence, $x$ is the largest integer divisor of 999 less than 20 i.e. $\boxed{9}$