To solve the problem of finding the sum of all positive integers less than 1000 that are divisible by 3 or 5, we use the principle of inclusion-exclusion:

Step 1: Sum of numbers divisible by 3

Numbers: 3, 6, ..., 999
Number of terms: $\lfloor \frac{999}{3} \rfloor = 333$
Sum: $\frac{333(3 + 999)}{2} = 333 \times 501 = 166833$

Step 2: Sum of numbers divisible by 5

Numbers: 5, 10, ..., 995
Number of terms: $\lfloor \frac{999}{5} \rfloor = 199$
Sum: $\frac{199(5 + 995)}{2} = 199 \times 500 = 99500$

Step 3: Sum of numbers divisible by both 3 and 5 (i.e., 15)

Numbers: 15, 30, ..., 990
Number of terms: $\lfloor \frac{999}{15} \rfloor = 66$
Sum: $\frac{66(15 + 990)}{2} = 33 \times 1005 = 33165$

Step 4: Total sum (inclusion-exclusion)

Total = Sum(3) + Sum(5) - Sum(15)
Total = $166833 + 99500 - 33165 = 233168$

Answer: $\boxed{233168}$

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