6・119 Every natural number $n$ greater than 2 can be expressed as the sum of several pairwise distinct natural numbers. Let the maximum number of terms be $A(n)$. Find $A(n)$ (expressed in terms of $n$).
[Solution] Let $n=a_{1}+a_{2}+\cdots+a_{A(n)}, a_{1}<a_{2}<\cdots<a_{A(n)}$ be natural numbers.

Obviously, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{A(n)} \geqslant A(n)$, therefore, $n \geqslant 1+2+\cdots+$ $A(n)=\frac{1}{2} A(n)(A(n)+1)$,
thus, the natural number $A(n) \leqslant\left[\frac{\sqrt{8 n+1}-1}{2}\right]$.
$$\text { Also } \begin{aligned}
n= & 1+2+\cdots+\left(\left[\frac{\sqrt{8 n+1}-1}{2}\right]-1\right) \\
& +\left(\left[\frac{\sqrt{8 n+1}-1}{2}\right]+a\right)
\end{aligned}$$

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

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