How many solutions does the inequality

$$
|x|+|y|<1000
$$

have, where $x$ and $y$ are integers?
I. Solution: The value of each unknown can only be one of the numbers $0, \pm 1, \pm 2, \ldots, \pm 999$, but the choice of one value imposes a restriction on the other. Let's go through each value of $x$ and determine the number of possible $y$ values that solve the equation with it; the sum of these numbers will give us the answer to our question. - First, let's consider the non-negative values of $x$.

For $x=0$, $|y|<1000$, i.e., $-1000<y<1000$, which is satisfied by $-999, -998, \ldots, -1, 0, 1, \ldots, 999$, the number of which is $2 \cdot 999 + 1 = 1999$.

For $x=1$, $|y|<999$, there are 2 fewer solutions than before, because $y = \pm 999$ no longer fits.

In general, when transitioning from a certain value of $x$ to the next higher one, the largest possible value of $|y|$ decreases by 1, so we need to remove two of the previous values of $y$, and the number of solutions for $x, y$ decreases by 2. Thus, the number of solutions for $x=0, 1, 2, \ldots, 999$ forms an arithmetic sequence. The last term for $x=999$ is 1, the number of terms is 1000, and thus their sum is:

$$
\frac{1000(1999+1)}{2} = 1000^2
$$

For $x=-1, -2, \ldots, -999$, $|x|=1, 2, \ldots, 999$, similarly, the total number of such solutions is

$$
1997 + 1995 + \ldots + 1 = 999^2
$$

With this, we have considered all solutions, and their total number is: $1000^2 + 999^2 = 1998001$.

II. Solution: The question is clear with the following: how many pairs of integers $x, y$ are there such that the sum $|x| + |y|$ is either 0, 1, 2, ..., or 999? Therefore, let's determine the number $N_k$ of integer solutions to the equation $|x| + |y| = k$, where $k$ is a non-negative integer, and sum these for $k=0, 1, 2, \ldots, 999$.

For $k=0$, there is one solution: $x = y = 0$.

For $k=1$, there are four suitable value pairs:

1, 0; -1, 0; 0, 1; 0, -1.

For $k \geq 2$, we divide the solutions into two groups depending on whether the number 0 appears in them or not. For the first group, similar to the case $k=1$, there are 4 solutions: $k, 0; -k, 0; 0, k; 0, -k$. For the second group, $|x|$ can take the following $k-1$ values:

$$
|x| = j = 1, 2, 3, \ldots, k-1
$$

and with this, $|y|$ is also determined:

$$
|y| = k - j = k - 1, k - 2, k - 3, \ldots, 1
$$

For each value of $j$, considering the signs, 4 solutions arise, because $j$ and $k-j$ can each independently take 2 signs:

$$
j, k-j; -j, k-j; j, -(k-j); -j, -(k-j)
$$

Thus, the number of solutions in this case is $4(k-1)$. - In total, for $k \geq 2$, the number of solutions is $4 + 4(k-1) = 4k$. This expression is also valid for $k=1$, in which case the second group does not contain solutions; however, it is not valid for $k=0$.

After this, for $k=0, 1, 2, \ldots, 999$:

$$
N = 1 + 4(1 + 2 + 3 + \ldots + 999) = 1 + 4 \cdot \frac{999 \cdot 1000}{2} = 1998001
$$

Remark. Similarly, if $s$ is a non-negative integer, then the number of solutions to the inequality $|x| + |y| < s$ in integers is:

$$
N_s = 1 + 4[1 + 2 + 3 + \ldots + (s-1)] = 2s^2 - 2s + 1
$$

III. Solution: Each solution is represented in a rectangular coordinate system by a so-called lattice point, i.e., a point whose both coordinates are integers. Let's characterize the position of these lattice points, then determine their number and answer the question.

