1. **Problem statement:** We have balls numbered from 1 to 2020 in a box. We draw balls without replacement. We want to find the minimum number of balls drawn to guarantee that among them there are four balls with consecutive numbers. 2. **Understanding the problem:** We want to ensure that no matter how the balls are drawn, there will be at least one set of four consecutive numbers among the drawn balls. 3. **Key idea:** Use the pigeonhole principle and consider the worst-case scenario where we try to avoid having 4 consecutive numbers. 4. **Approach:** To avoid having 4 consecutive numbers, the drawn balls can have at most 3 consecutive numbers in a row. So, we want to find the maximum number of balls we can draw without having 4 consecutive numbers. 5. **Constructing the maximum set without 4 consecutive numbers:** - We can have groups of at most 3 consecutive numbers separated by at least one number not drawn. - Each group of 3 consecutive numbers is followed by at least one number skipped to prevent 4 consecutive. 6. **Counting the maximum number of balls without 4 consecutive:** - Each block consists of 3 drawn balls + 1 skipped ball = 4 numbers. - Number of such blocks in 2020 numbers is $\left\lfloor \frac{2020}{4} \right\rfloor = 505$ blocks. - Each block contributes 3 drawn balls, so total drawn balls without 4 consecutive is $3 \times 505 = 1515$. - Remaining numbers after 505 blocks: $2020 - 4 \times 505 = 2020 - 2020 = 0$. 7. **Conclusion:** - Maximum balls drawn without 4 consecutive is 1515. - Therefore, to guarantee 4 consecutive balls, we need to draw at least $1515 + 1 = 1516$ balls. **Final answer:** $$\boxed{1516}$$