24. How many integer solutions does the inequality

$$
|x|+|y|<100 ?
$$

have? Here, for $x \neq y$, the solutions $x, y$ and $y, x$ should be considered different.
24. We need to find the number of pairs of integers \(x, y\) such that \(|x| + |y|\) is 0, 1, 2, 3, ..., or 99. Let's count the number of pairs \(x, y\) for which \(|x| + |y| = k\). Here, \(|x|\) can take \(k + 1\) different values, namely \(0, 1, 2, \ldots, k-1, k\); in this case, \(|y|\) will be \(k, k-1, k-2, \ldots, 1, 0\). If \(|x| = l\) and \(|y| = k - l\), where neither \(l\) nor \(k - l\) is 0, then \(x\) and \(y\) can each take two values differing in sign; thus, we get four different solutions to the equation \(|x| + |y| = k\), namely: \((l, k-l)\),
\((-l, k-l)\), \((l, -k+l)\) and \((-l, k-l)\). If \(l = k\) or \(l = 0\), then we get only two solutions to the equation \(|x| + |y| = k\), namely: \((0, k)\) and \((0, -k)\) or \((k, 0)\) and \((-k, 0)\). Therefore, for \(k \neq 0\), the equation \(|x| + |y| = k\) has \(2 + 4(k-1) + 2 = 4k\) different solutions. If \(k = 0\), then the equation \(|x| + |y| = k\) (i.e., the equation \(|x| + |y| = 0\)) has one unique solution: \(x = 0, y = 0\). Therefore, the number of different integer solutions to the inequality \(|x| + |y| < 100\) is \(1 + 4(1 + 2 + 3 + \ldots + 99) = 1 + 4 \cdot \frac{99 \cdot 100}{2} =\)

\[
= 1 + 19800 = 19801
\]