Answer: 400. Solution. Note that for any integers $a, b$ the system of equations $\left\lvert\,\left\{\begin{array}{l}3 x+2 y=a \\ 2 x+y=b\end{array}\right.$ has an integer solution $\left\lvert\,\left\{\begin{array}{l}x=2 b-a \\ y=2 a-3 b\end{array}\right.$, and different ordered pairs $(a, b)$ correspond to different solutions ( $x, y$ ). Therefore, for any natural number $n$ from 1 to 99, taking $a, b$ as numbers of the form $a= \pm n, b= \pm(100-n)$, we get $99 \cdot 4=396$ solutions. In addition, there will be two solutions corresponding to the numbers $a= \pm 100, b=0$ and, similarly, there will be two more solutions corresponding to the numbers $a=0, b= \pm 100$. In total, 400 solutions.
10.4. How many solutions in integers $x, y$ does the equation $|3 x+2 y|+|2 x+y|=100$ have?