17. If $a$ is a number that satisfies the condition, then $a^{x_{0}}>1$, so $a>1$. At this point, taking

both sides modulo $a-1$, we know
so $\square$
$$\begin{array}{c} a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}} \\ 1 \equiv \underbrace{1+\cdots+1}_{2001 \uparrow}(\bmod a-1), \\ a-1 \mid 2000 . \end{array}$$

On the other hand, if $a>1$ and $a-1 \mid 2000$, then we can take $x_{1}, x_{2}, \cdots, x_{2001}$ to be $a$ numbers as $0$, $a-1$ numbers as $1$, $a-1$ numbers as $2$, ..., $a-1$ numbers as $k-1$, where $k=\frac{2000}{a-1}$, and take $x_{0}=k$, then we have $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$.

Therefore, $a$ is a number that satisfies the condition if and only if $a>1$ and $a-1 \mid 2000$, and there are 20 such $a$.