Solution:

For some $i$, $g_{i}(1999)$ is odd. Indeed, if this were false, each $g_{i}(1999)$ would be divisible by $2$, so their product, $f(1999) = 2000$, would be divisible by $2^{5}$, which is not the case.

Now write $g_{i}(x) = c_{n} x^{n} + c_{n-1} x^{n-1} + \cdots + c_{0}$. Then
$$g_{i}(1999) - (c_{n} + \cdots + c_{0}) = c_{n}(1999^{n} - 1) + c_{n-1}(1999^{n-1} - 1) + \cdots + c_{0}(1 - 1),$$
which is even since $1999^{k} - 1$ is even for each $k$. Since $g_{i}(1999)$ is odd, $c_{n} + \cdots + c_{0}$ is also odd, as desired.