［Solution］ After calculation, we can obtain
$$\begin{aligned} a_{1} & =\frac{a \sqrt{3}+1}{\sqrt{3}-a} \\ a_{2} & =\frac{a_{1} \sqrt{3}+1}{\sqrt{3}-a_{1}}=\frac{(a \sqrt{3}+1) \sqrt{3}+(\sqrt{3}-a)}{\sqrt{3}(\sqrt{3}-a)-(a \sqrt{3}+1)} \\ & =\frac{a+\sqrt{3}}{1-a \sqrt{3}} \\ a_{3} & =\frac{a_{2} \sqrt{3}+1}{\sqrt{3}-a_{2}}=\frac{(a+\sqrt{3}) \sqrt{3}+(1-a \sqrt{3})}{\sqrt{3}(1-a \sqrt{3})-(a+\sqrt{3})} \\ & =-\frac{1}{a} \end{aligned}$$

From this, we get $\square$
$$a_{6}=-\frac{1}{-\frac{1}{a}}=a$$

It is evident that the sequence $a_{0}, a_{1}, a_{2}, \cdots$ is a periodic sequence with a period of 6.
Since $1994=6 \times 332+2$, we have
$$a_{1994}=a_{2}=\frac{a+\sqrt{3}}{1-a \sqrt{3}} .$$

Where $k, u$ are both positive integers.
Substituting $k=u u_{5}$, we get
$$\begin{array}{l} u^{2} \cdot u_{5}=2000 \\ u^{2} \cdot u_{5}=2^{4} \cdot 5^{3} \end{array}$$
All possible values of $u^{2}$ are $1,2^{2}, 2^{4}, 5^{2}, 2^{2} \cdot 5^{2}, 2^{4} 5^{2}$, thus all possible values of $u$ are 1, $2,2^{2}, 5,2 \cdot 5,2^{2} \cdot 5$; all possible values of $k$ are $2000,1000,500,400,200,100$.