Let $\lambda$ the positive root of the equation $t^2-1998t-1=0$.  It is defined the sequence $x_0,x_1,x_2,\ldots,x_n,\ldots$ by $x_0=1,\ x_{n+1}=\lfloor\lambda{x_n}\rfloor\mbox{ for }n=1,2\ldots$ Find the remainder of the division of $x_{1998}$ by $1998$. 

Note: $\lfloor{x}\rfloor$ is the greatest integer less than or equal to $x$.
1. **Find the positive root of the quadratic equation:**
   The given equation is \( t^2 - 1998t - 1 = 0 \). To find the roots, we use the quadratic formula:
   \[
   t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 1 \), \( b = -1998 \), and \( c = -1 \). Plugging in these values, we get:
   \[
   t = \frac{1998 \pm \sqrt{1998^2 + 4}}{2}
   \]
   Simplifying further:
   \[
   t = \frac{1998 \pm \sqrt{3992004}}{2}
   \]
   \[
   t = \frac{1998 \pm 1998.001}{2}
   \]
   The positive root is:
   \[
   \lambda = \frac{1998 + 1998.001}{2} \approx 1998.0005
   \]

2. **Define the sequence and derive the recurrence relation:**
   The sequence is defined by \( x_0 = 1 \) and \( x_{n+1} = \lfloor \lambda x_n \rfloor \). We need to find the behavior of this sequence modulo 1998.

3. **Analyze the sequence modulo 1998:**
   We have:
   \[
   x_{n+1} = \lfloor \lambda x_n \rfloor
   \]
   Given \( \lambda \approx 1998 \), we can write:
   \[
   x_{n+1} \approx 1998 x_n
   \]
   Since we are interested in the sequence modulo 1998, we can simplify:
   \[
   x_{n+1} \equiv \lfloor 1998 x_n \rfloor \pmod{1998}
   \]
   Notice that:
   \[
   \lfloor 1998 x_n \rfloor \equiv 0 \pmod{1998}
   \]
   This implies:
   \[
   x_{n+1} \equiv 0 \pmod{1998}
   \]

4. **Establish the periodicity and find \( x_{1998} \):**
   From the recurrence relation, we see that \( x_{n+1} \equiv x_{n-1} - 1 \pmod{1998} \). This suggests a periodic behavior. Specifically:
   \[
   x_{n+1} \equiv x_{n-1} - 1 \pmod{1998}
   \]
   Given \( x_0 = 1 \), we can compute the first few terms to identify the pattern:
   \[
   x_1 = \lfloor \lambda \cdot 1 \rfloor = 1998
   \]
   \[
   x_2 = \lfloor \lambda \cdot 1998 \rfloor \equiv 0 \pmod{1998}
   \]
   \[
   x_3 = \lfloor \lambda \cdot 0 \rfloor = 0
   \]
   \[
   x_4 = \lfloor \lambda \cdot 0 \rfloor = 0
   \]
   From this, we see that \( x_n \equiv 0 \pmod{1998} \) for \( n \geq 2 \).

5. **Conclusion:**
   Since \( x_n \equiv 0 \pmod{1998} \) for \( n \geq 2 \), it follows that:
   \[
   x_{1998} \equiv 0 \pmod{1998}
   \]

The final answer is \( \boxed{0} \)