4. Divide 1996 into the sum of 96 positive integers, the maximum value of the product of these 96 positive integers is
4. $20^{20} \cdot 21^{76}$

Let $1996=a_{1}+a_{2}+\cdots+a_{96}$, and $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{96}$.
If $a_{06} \geqslant a_{1}+2$, then let $a^{\prime}{ }_{06}=a_{06}-1, a_{1}^{\prime}=a_{1}+1$, so $a_{96}^{\prime}+a_{1}^{\prime}=a_{96}+a_{1}$, and
$$
a_{1}^{\prime} \cdot a_{96}^{\prime}=\left(a_{1}+1\right)\left(a_{96}-1\right)>a_{1} a_{96},
$$

Therefore, by replacing $a_{1}, a_{86}$ with $a_{1}^{\prime}, a^{\prime}{ }_{06}$, the product of the 96 numbers increases.

Thus, it must satisfy $a_{96} \leqslant a_{1}+1$. Since $a_{06} \geqslant a_{1}$, then $a_{06}=$ $a_{1}$ or $a_{96}=a_{1}+1$. Given $96 \times 1996$, so $a_{06}=a_{1}+\mathrm{i}$.
Because $1996=20 \times 96+76=30 \times 20+21 \times 76$,
we can divide 1996 into 96 numbers, 20 of which are 20 and 76 of which are 21, so the product of these 96 numbers is maximized, with the maximum value being $20^{20} \cdot 21^{76}$.