jmc

From the formula for a geometric series, $$704+\frac{704}{2}+\frac{704}{4}+\cdots +\frac{704}{2^{n-1}}=704\cdot \frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}=1408\left(1-\frac{1}{2^n}\right),$$and $$1984-\frac{1984}{2}+\frac{1984}{4}+\cdots +\frac{1984}{(-2{)}^{n-1}}=1984\cdot \frac{1-\frac{1}{(-2{)}^n}}{1+\frac{1}{2}}=\frac{3968}{3}\left(1-\frac{1}{(-2{)}^n}\right).$$Hence, $$1408\left(1-\frac{1}{2^n}\right)=\frac{3968}{3}\left(1-\frac{1}{(-2{)}^n}\right).$$This reduces to $$33\left(1-\frac{1}{2^n}\right)=31\left(1-\frac{1}{(-2{)}^n}\right).$$If $n$ is even, then $(-2{)}^n=2^n,$ and there are no solutions. Otherwise, $n$ is odd, and $(-2{)}^n=-2^n,$ so $$33\left(1-\frac{1}{2^n}\right)=31\left(1+\frac{1}{2^n}\right).$$Isolating $2^n,$ we get $2^n=32,$ so $n=\boxed{5}.$