To find the number of ways to write the number $2000$ as a sum of three positive integers $a, b, c$ such that $a \leq b \leq c$, we can follow these steps:

1. Define the problem in terms of inequalities:
We need to find the number of solutions to the equation $a + b + c = 2000$ where $a, b, c \in \mathbb{N}^*$ and $a \leq b \leq c$.

2. **Determine the range for $a$:**
Since $a \leq b \leq c$, the smallest value $a$ can take is $1$. The largest value $a$ can take is $\left\lfloor \frac{2000}{3} \right\rfloor = 666$.

3. **Determine the range for $b$ given $a$:**
For a given $a$, $b$ must satisfy $a \leq b \leq \frac{2000 - a}{2}$. This is because $b$ must be at least $a$ and at most $\frac{2000 - a}{2}$ to ensure $b \leq c$.

4. **Calculate the number of possible values for $b$:**
- The minimum value of $b$ is $b_1 = a$.
- The maximum value of $b$ is $b_2 = \left\lfloor \frac{2000 - a}{2} \right\rfloor$.

5. **Count the number of possible values for $b$:**
The number of possible values for $b$ for a given $a$ is $N_a = b_2 - b_1 + 1$.

6. **Express $N_a$ in terms of $a$:**
- For $a$ even, let $a = 2m$:
$$N_a = \left\lfloor \frac{2000 - 2m}{2} \right\rfloor - 2m + 1 = 1000 - m - 2m + 1 = 1001 - 3m$$
- For $a$ odd, let $a = 2m - 1$:
$$N_a = \left\lfloor \frac{2000 - (2m - 1)}{2} \right\rfloor - (2m - 1) + 1 = 1000 - m + 1 - 2m + 1 = 1002 - 3m$$

7. **Sum the number of triplets for all possible values of $a$:**
$$N = \sum_{m=1}^{333} (1001 - 3m) + \sum_{m=1}^{333} (1002 - 3m)$$
$$N = \sum_{m=1}^{333} (2003 - 6m)$$

8. Calculate the sum:
$$N = \sum_{m=1}^{333} 2003 - \sum_{m=1}^{333} 6m$$
$$N = 333 \times 2003 - 6 \sum_{m=1}^{333} m$$
The sum of the first 333 natural numbers is:
$$\sum_{m=1}^{333} m = \frac{333 \times 334}{2} = 55506$$
Therefore:
$$N = 333 \times 2003 - 6 \times 55506$$
$$N = 666999 - 333036 = 333963$$

The final answer is $\boxed{333963}$.