Group the numbers into $7$ groups according to their remainders upon division by $7$: the groups are
$$\begin{aligned} \{1, 8, 15, \dots, 498\}, & \{2, 9, 16, \dots, 499\}, \{3, 10, 17, \dots, 500\}, \\ \{4, 11, 18, \dots, 494\}, & \{5, 12, 19, \dots, 495\}, \{6, 13, 20, \dots, 496\}, \{7, 14, 21, \dots, 497\}. \end{aligned}$$
Note that if a number from a group that leaves remainder $x$ is not deleted, then all the numbers from the group that leaves remainder $7-x$ must be deleted, otherwise two numbers will sum to a multiple of $7$. In particular, only one number from the last group above may be selected.
Note also that if one number from a group (besides the last one) is selected, we may select all numbers in that group, since they all leave the same remainder when divided by $7$. The first three groups contain $72$ numbers each, while the last four contain $71$ numbers each. If we select the first three groups, we may not select any number in the next three groups, and finally we can select one number from the last group. This yields $72 \times 3 + 1 = 217$ numbers.