4. In $\pm 1 \pm 2 \pm 3 \pm 5 \pm 20$, by appropriately choosing + or -, different algebraic sums can be obtained $\qquad$.


4.24.

Among $1,2,3,5,20$, there are three odd numbers, so their algebraic sum must be odd.

Observing, we see that from $1,2,3,5$ we can obtain all odd numbers with absolute values not exceeding 11.

According to the problem, the expression must include 1, 2, 3, 5, and 20. Therefore, the integers that can be obtained are $\pm 20$ plus or minus $1,3,5,7,9, 11$, i.e., 12 positive odd numbers 9, $11, \cdots, 31$ and 12 negative odd numbers $-9,-11, \cdots$, -31.
Thus, the total number of integers that can be expressed is 24.