8. In the sequence $\left\{a_{n}\right\}_{n=1}^{+\infty}$, $x_{1}=1$, if for any positive integer $n$ there is $x_{n+1}=x_{n}+\left[\sqrt{x_{n}}\right]$, where the symbol $[x]$ denotes the greatest integer not exceeding $x$, then the smallest positive integer $n$ such that $x_{n} \geqslant 2019$ is $\qquad$.
$$
\begin{array}{l}
x_{1}=1, x_{2}=1+1=2, x_{3}=2+1=3, x_{4}=3+1=4, x_{5}=4+2=6, \\
x_{6}=6+2=8, x_{7}=8+2=10, x_{8}=10+3=13, x_{9}=13+3=16, \\
x_{10}=16+4=20, x_{11}=20+4=24, x_{12}=24+4=28, x_{13}=28+5=33, \\
x_{14}=33+5=38, x_{15}=38+6=44, x_{16}=44+6=50, x_{17}=50+7=57, \\
x_{18}=57+7=64, x_{19}=64+8=72, x_{20}=72+8=80, \cdots .
\end{array}
$$

Assume $x_{k}=m^{2}$, then $x_{k+1}=m^{2}+m, x_{k+2}=m^{2}+2 m, x_{k+3}=(m+1)^{2}+m-1$,
$$
x_{k+4}=(m+1)^{2}+2 m, x_{k+5}=(m+2)^{2}+m-2, \cdots, x_{k+2 m+1}=4 m^{2} \text {. }
$$

Thus, $x_{1}=1^{2}, x_{1+3}=2^{2}, x_{4+5}=4^{2}, x_{9+9}=8^{2}, x_{18+17}=16^{2}, x_{35+33}=32^{2}, \cdots$,
$$
1+3+5+9+17+\cdots+2^{n-1}+1=2^{n}+n-2 \Rightarrow x_{2^{n}+n-2}=2^{2 n-2}\left(n \in \mathbf{N}^{*}\right) \text {. }
$$

Since $2019>1024=2^{10}$, and $x_{64+6-2}=x_{68}=32^{2}$,
then $x_{68+2 r+1}=(32+r)^{2}+32-r \Rightarrow x_{68+25}=44^{2}+20=1956$,
$$
x_{94}=1956+44=2000, x_{95}=2000+44=2044 \text {, }
$$

Therefore, the smallest positive integer $n$ such that $x_{n} \geqslant 2019$ is 95.