Let $a,b,c\in\mathbb{N}^*, a\le b\le c, a+b+c=2000$. Results $1\le a\le \left\lfloor\dfrac{2000}{3}\right\rfloor=666$. For a given number $a$, $a\le b\le \dfrac{2000-a}{2}$. The minimum value of $b$ is $b_1=a$ and the maximum value of $b$ is: $b_2=1000-\dfrac{a}{2}$ for $a$ even number, respectively $b_2=1000-\dfrac{a+1}{2}$ for $a$ odd number. Hence, for a given number $a$, exist $N_a=b_2-b_1+1$ possible values for $b$ and automatically $N_a$ distinct triplets $(a,b,c)$. For $a$ even number denote $a=2m$ and for $a$ odd number denote $a=2m-1$. For $a=2m: N_a=1000-\dfrac{a}{2}-a+1=1001-\dfrac{3a}{2}=1001-3m$. For $a=2m-1: N_a=1000-\dfrac{a+1}{2}-a+1=1001-\dfrac{3a+1}{2}=1002-3m$. The total number of the triplets with the property $a,b,c\in\mathbb{N}^*, a\le b\le c, a+b+c=2000$ is: $N=\sum_{a=1}^{666}N_a=\sum_{m=1}^{333}(1001-3m)+\sum_{m=1}^{333}(1002-3m)=\sum_{m=1}^{333}(2003-6m)=333333$.