Solution: We transform the expression by multiplying the numerator and denominator of each fraction by the expression conjugate to the denominator:

$$
\begin{aligned}
& A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}= \\
& =\frac{\sqrt{a_{2}}-\sqrt{a_{1}}}{a_{2}-a_{1}}+\frac{\sqrt{a_{3}}-\sqrt{a_{2}}}{a_{3}-a_{2}}+\ldots+\frac{\sqrt{a_{1580}}-\sqrt{a_{1579}}}{a_{1580}-a_{1579}}=
\end{aligned}
$$

$=\frac{\sqrt{a_{2}}-\sqrt{a_{1}}}{d}+\frac{\sqrt{a_{3}}-\sqrt{a_{2}}}{d}+\ldots+\frac{\sqrt{a_{1580}}-\sqrt{a_{1579}}}{d}=$

$=\frac{\sqrt{a_{1580}}-\sqrt{a_{1}}}{d}=\frac{\sqrt{a_{1}+1579 d}-\sqrt{a_{1}}}{d}=\frac{\sqrt{1+1579 \cdot 4}-\sqrt{1}}{4}=\frac{\sqrt{6317}-1}{4}$

We estimate the value of $A: \quad 19<\frac{\sqrt{6317}-1}{4}<20$, and write the smallest integer greater than $A$ in the answer.

Answer: 20.
3. In the arithmetic progression $\left(a_{n}\right) a_{1}=1, d=4$.

$$
\text { Calculate } A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}
$$

In the answer, write the smallest integer greater than $A$.