Example 3 If $a_{1}=1, a_{k}=k+a_{k-1}(2 \leqslant k \leqslant$ $n)$, then $a_{1}, a_{2}, \cdots, a_{n}$ is called a regular number. Question: How many numbers in the set $\{1,2, \cdots, 2001\}$ are either a regular number itself or the sum of several different regular numbers?
Solution: It is easy to know that $a_{k}=\frac{k(k+1)}{2}=C_{k+1}^{2}$, these regular numbers form the sequence $1,3,6,10,15,21,28,36,45,55$, ... First, we check the numbers that can be represented by regular numbers (i.e., can be written as the sum of several different regular numbers):

Among the positive integers less than 6, 2 and 5 cannot be represented, while the others can be represented by regular numbers less than 6;

Among the positive integers less than 10, 2, 5, and 8 cannot be represented, while the others can be represented by regular numbers less than 10;

Among the positive integers less than 15, $2,5,8,12$ cannot be represented, while the others can be represented by regular numbers less than 15;

Among the positive integers less than 21, $2,5,8,12$ cannot be represented, while the others can be represented by regular numbers less than 21;

Among the positive integers less than 28, $2,5,8,12,23$ cannot be represented, while the others can be represented by regular numbers less than 28;

Among the positive integers less than 36, $2,5,8,12,23,33$ cannot be represented, while the others can be represented by regular numbers less than 36;

Among the positive integers less than 45, $2,5,8,12,23,33$ cannot be represented, while the others can be represented by regular numbers less than 45;

Among the positive integers less than 55, $2,5,8,12,23,33$ cannot be represented, while the others can be represented by regular numbers less than 55;

Among the positive integers less than 66, $2,5,8,12,23,33$ cannot be represented, while the others can be represented by regular numbers less than 66.

By checking more groups, we can be confident that:
When $n \geqslant 36$, in the set $\{1,2, \cdots, n\}$, the non-regular numbers that cannot be represented by regular numbers are only 6, which are 2, $5,8,12,23,33$.
Below, we use mathematical induction to prove it.
Assume that for all positive integers less than $\mathrm{C}_{k+1}^{2}(k \geqslant 11)$, except for $2,5,8,12,23,33$, all other non-regular numbers can be represented by regular numbers. We only need to prove that all integers $x$ between $\mathrm{C}_{k+1}^{2}$ and $\mathrm{C}_{k+2}^{2}$ can be represented by regular numbers.
(1) If $x-C_{k+1}^{2}$ is not one of $2,5,8,12,23,33$, then $x-\mathrm{C}_{k+1}^{2}$ can be represented by regular numbers less than $\mathrm{C}_{k+1}^{2}$. Clearly, $x$ can be represented by regular numbers.
(2) If $x-\mathrm{C}_{k+1}^{2}=2$, then $x-\mathrm{C}_{k}^{2}=k+2 \geqslant$ 13. When $k \neq 21$ and $k \neq 31$, since $x-C_{k}^{2}=k+2$ $<\mathrm{C}_{k}^{2}$, then $x-\mathrm{C}_{k}^{2}$ can be represented by regular numbers less than $C_{k}^{2}$, thus, $x$ can be represented by regular numbers; When $k=21$ or 31, $x-C_{k-1}^{2}=2 k+1=43$ or 63, which can clearly be represented by regular numbers less than $\mathrm{C}_{k-1}^{2}$ (43=28+15; 63=28+21+10+3+1), thus, $x$ can be represented by regular numbers.
(3) If $x-\mathrm{C}_{k+1}^{2}=5$, then $x-\mathrm{C}_{k}^{2}=k+5 \geqslant$ 16. When $k \neq 18$ and $k \neq 28$, $x-\mathrm{C}_{k}^{2}$ can be represented by regular numbers less than $\mathrm{C}_{k}^{2}$, thus, $x$ can be represented by regular numbers; When $k=18$ or 28, $x-C_{k-1}^{2}=2 k+4=40$ or 60, which can clearly be represented by regular numbers less than $\mathrm{C}_{k-1}^{2}$ (40=21+15+3+1; 60=28+21+10+1), thus, $x$ can be represented by regular numbers.
(4) If $x-\mathrm{C}_{k+1}^{2}=8$, then $x-\mathrm{C}_{k}^{2}=k+8 \geqslant$ 19. When $k \neq 15$ and $k \neq 25$, $x-C_{k}^{2}$ can be represented by regular numbers less than $\mathrm{C}_{k}^{2}$, thus, $x$ can be represented by regular numbers; When $k=15$ or 25, $x-\mathrm{C}_{k-1}^{2}=2 k+7=37$ or 57, which can clearly be represented by regular numbers less than $\mathrm{C}_{k-1}^{2}$ (37=21+15+1; 57=28+15+10+3+1), thus, $x$ can be represented by regular numbers.
(5) If $x-\mathrm{C}_{k+1}^{2}=12$, then $x-\mathrm{C}_{k}^{2}=k+12 \geqslant$ 23. When $k \neq 11$ and $k \neq 21$, $x-C_{k}^{2}$ can be represented by regular numbers less than $\mathrm{C}_{k}^{2}$, thus, $x$ can be represented by regular numbers; When $k=21$, $x-\mathrm{C}_{k-1}^{2}=2 k+11=53$, which can clearly be represented by regular numbers less than $\mathrm{C}_{k-1}^{2}$ (53=28+15+10), thus, $x$ can be represented by regular numbers; When $k=11$, $x-C_{k-2}^{2}=3 k+9=42$, which can clearly be represented by regular numbers less than $C_{k-2}^{2}$ (42=28+10+3+1), thus, $x$ can be represented by regular numbers.
(6) If $x-\mathrm{C}_{k+1}^{2}=23$ or $x-\mathrm{C}_{k+1}^{2}=33$, then $x-\mathrm{C}_{k}^{2}=k+23$ or $x-\mathrm{C}_{k}^{2}=k+33$, both of which are greater than 33 (since $k \geqslant 11$), so $x-\mathrm{C}_{k}^{2}$ can be represented by regular numbers less than $\mathrm{C}_{k}^{2}$. Thus, $x$ can be represented by regular numbers.

In summary, in the set $\{1,2, \cdots, 2001\}$, the non-regular numbers that cannot be represented by regular numbers are only 6, which are $2,5,8,12,23,33$. Therefore, the number of numbers that are regular numbers themselves or can be represented by regular numbers is 1995.