Can 2123456 be split into the sum of
(1) 64; (2) 128

consecutive natural numbers?
Solution: (1) When $r=64$, since $\frac{2 n}{r}=2 \times 123456 \div 64=3858$ has the same parity as $r=64$, 123456 cannot be decomposed into the sum of 64 consecutive natural numbers; $\square$

(2) When $r=128$, since $\frac{2 n}{r}=2 \times 123456 \div 128=1929$ has a different parity from $r=64$, and $2 \times 123456 \geqslant 1282 \times (128+1)$, 123456 can be decomposed into the sum of 128 consecutive natural numbers. In fact,
$$
123456=901+902+\cdots+1028.
$$