4. Divide the 20 positive integers from $1 \sim 20$ into two groups $A$ and $B$, such that the sum of all numbers in group $A$ equals $N$, and the product of all numbers in group $B$ also equals $N$. Then all possible values of $N$ are $\qquad$
4.180, 182, 192.

Let the numbers in group $B$ be $a, b, c, d, \cdots$, and the numbers are in ascending order according to their letter sequence. Then,
$$
N=a b c d \cdots=210-(a+b+c+d+\cdots) \text {. }
$$

Therefore, if $N5$, then
$$
N>1 \times 2 \times 3 \times 4 \times 5 \times 6=720 \text {, }
$$

which contradicts $N1$. When $I=2$, if $a=1$, then $b=2$, and $(6 c+1)(6 d+1)=1231$, which has no solution.

When $a=1, b=2,3,4$, similar discussions for the three cases show that there is no $N$ that satisfies the condition.
(4) If $I=5$, since $1 \times 2 \times 4 \times 5 \times 6>210$, then $a=1, b=2, c=3$. Hence $d e3$, and $(d, e)=(4,5), (4,6), (4,7), (4,8), (5,6)$.
Upon inspection, only $(4,8)$ satisfies the condition, so $N=192$. Therefore, all possible values of $N$ are $180, 182, 192$.