We broke down 1996 into the sum of some integers. What remainder does the sum of the cubes of these numbers give when divided by 6?
For any integer $a$, $a^{3}-a=(a-1) a(a+1)$ is divisible by 6, because one of the three factors on the right side is divisible by 3, and among the three factors, there is certainly an even number.

Therefore, if $1996=a_{1}+a_{2}+\cdots+a_{n}$, then

$$
a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}-1996=\left(a_{1}^{3}-a_{1}\right)+\left(a_{2}^{3}-a_{2}\right)+\cdots+\left(a_{n}^{3}-a_{n}\right)
$$

is certainly divisible by 6. Thus, $a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}$, when divided by 6, gives the same remainder as 1996, which is 4.

Based on the work of Zsuzsanna Tóth 515 (Szekszárd, Garay J. Gymnasium, 1st year).