11. (5 points) In the 200 natural numbers from 1 to 200, how many numbers must be selected to ensure that there are at least 2 numbers whose product is 238?
【Analysis】First analyze the factors of 238, making 2 of the factors multiply to 238, there are 4 groups in total, and use the worst-case principle to find the result.

【Solution】Solution: According to the problem:
Decompose 238 into the product of numbers less than 200, we have $238=17 \times 14=7 \times 34=2 \times 119$, which gives us three groups.
The two-digit factors that multiply to 238 are $(17,14),(7,34),(2,119)$, totaling 6 numbers, which can be divided into 3 groups.
The worst-case principle is that the other 194 are chosen, then one is selected from each of the three groups, totaling 197. Selecting one more will ensure that the product of two numbers is 238.

Total $197+1=198$;
Answer: At least 198 must be selected to ensure that the product of two numbers is 238.