1. Obviously $a_{0}>a_{1}>a_{2}>\cdots$. Since $a_{k}-a_{k+1}=1-\frac{1}{a_{k}+1}$, we have $a_{n}=a_{0}+\left(a_{1}-a_{0}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=1994-n+\frac{1}{a_{0}+1}+\cdots+\frac{1}{a_{n-1}+1}>$ $1994-n$. Also, for $1 \leq n \leq 998$,
$$\frac{1}{a_{0}+1}+\cdots+\frac{1}{a_{n-1}+1} < 1.$$
Hence,
$$a_{n} < 1995 - n.$$
For $999 \leq n \leq 1993$,
$$\frac{1}{a_{0}+1}+\cdots+\frac{1}{a_{n-1}+1} > 1.$$
Thus,
$$a_{n} < 1994 - n.$$
For $n = 1994$,
$$\frac{1}{a_{0}+1}+\cdots+\frac{1}{a_{1993}+1} = 1.$$
Therefore,
$$a_{1994} = 0.$$
For $n \geq 1995$,
$$\frac{1}{a_{0}+1}+\cdots+\frac{1}{a_{n-1}+1} > 1.$$
Hence,
$$a_{n} < 1994 - n.$$
In summary, for $1 \leq n \leq 998$,
$$\left\lfloor a_{n} \right\rfloor = 1994 - n.$$