2. A natural number minus 69 is a perfect square, and this natural number plus 20 is still a perfect square. Then this natural number is $\qquad$ .
2.2005 .

Let this natural number be $x$. According to the problem, we have
$$
\left\{\begin{array}{l}
x-69=m^{2}, \\
x+20=n^{2},
\end{array}\right.
$$

where $m, n$ are both natural numbers.
Subtracting the two equations gives $n^{2}-m^{2}=89$, which is $(n-m)(n+m)=89$.
Since $n>m$, and 89 is a prime number, we have
$$
\left\{\begin{array}{l}
n-m=1, \\
n+m=89 .
\end{array}\right.
$$

Solving these, we get $n=45, x=2005$.