1. Every natural number $n$ greater than 2 can be expressed as the sum of several distinct natural numbers. Let the maximum number of such distinct natural numbers be $A(n)$, find $A(n)$ (expressed in terms of $n$).
1. Let $n=a_{1}+a_{2}+\cdots+a_{A(n)}, a_{1}<a_{2}<\cdots<a_{A(n)}$ be all natural numbers.
Obviously, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{a} \geqslant A_{G}$. Therefore,
$$
n \geqslant 1-1-2+\cdots+A(n):=\frac{1}{2} A(n)(A(n)+1) \text {. }
$$

Thus, the natural number $A(n) \leqslant\left[\frac{\sqrt{8 n+1}-1}{2}\right]$.
$$
\begin{array}{l}
\text { Also } n=1+2+\cdots+\left(\left[\frac{\sqrt{8 n+1}-1}{2}\right]-1\right) \\
+\left(\left[\frac{\sqrt{8 n+1}-1}{2}\right]+a\right) \text {. } \\
\end{array}
$$

where $a$ is an integer.
From (*) we know $a \geqslant 0$. Therefore,
$$
A(n) \geqslant\left[\frac{\sqrt{8 n+1}-1}{2}\right] \text {. }
$$

In summary, $A(n)=\left[\frac{\sqrt{8 n+1}-1}{2}\right]$.