Three, let $T_i$ be the $i$-th term of the regular sequence, then $T_i=\frac{1}{2} i(i+1), i \geqslant 1$, thus for $n \leqslant 33$, the $n$ that does not satisfy the condition ("invalid" numbers) is $n \in I=\{2,5,8,12,23,33\}$.

Step-by-step calculation shows that for $34 \leqslant n \leqslant 66$, there is always $n=\sum T_{1}, T_{1} \leqslant 66 ; n \leqslant 132$ when, there is always $n-\sum T_{1}, T_{\text {. }}$ $\leqslant 66 ; n \leqslant 223$ when, there is always $n=\sum T_{1}, T_{1} \leqslant 91 ; n \leqslant 413$ when, there is always $n=\sum T_{1}, T, \leqslant 190 ; n \leqslant$ 738 when, there is always $n=\sum T_{1}, T_{1} \leqslant 325 ; n \leqslant 1441$ when, there is always $n=\sum T_{1}, T_{1} \leqslant 1378 ; n=1378+$ $a, a \in\{34, \cdots, 1441\}$ when, there is $n \leqslant 2819$.

Therefore, there are $2001-6=1995$ valid numbers.
Three. (50 points) If a positive integer is regular itself and it is the sum of different regular numbers, then this number is called "valid". Question: How many "valid" numbers are there in the set $\{1,2,3, \cdots, 2001\}$? ("Regular" numbers are $a_{1}, a_{2}, \cdots, a_{1}$, $\cdots$, where $a_{1}=1, a_{k}=k+a_{k-1}$, for all $k \geqslant 2$).