Example 8 List the positive integers that satisfy the condition "contain at least one digit 0 and are multiples of 4" in ascending order: $20,40,60,80,100,104, \cdots$ Then the 158th number in this sequence is ( ).
(A) 2000
(B) 2004
(C)2 0008
(D) 2012
Explanation: In positive integers, the characteristic of being a multiple of 4 is that the last two digits are multiples of 4, with a total of 25 cases (including 7 cases with the digit 0: $00,04,08,20,40,60$, 80; and 18 cases without 0: $12,16,24,28,32$, $36,44,48,52,56,64,68,72,76,84,88,92$, 96).

The problem requires "at least one digit 0", so the 25 cases should be classified based on whether they include 0.

Obviously, there are only 4 two-digit numbers that meet the condition, namely $20,40,60,80$; there are $9 \times 7=$ 63 three-digit numbers that meet the condition, i.e., $100,104, \cdots, 980$.

For four-digit numbers that meet the condition and have 1 as the thousandth digit, there are two scenarios:
(1) The hundredth digit is 0, and the tenth and unit digits are not 0, which results in 18 cases, i.e., 1012, $1016, \cdots, 1096$.
(2) The tenth and unit digits include the digit 0, in which case the hundredth digit can be any of $0 \sim 9$, so there are $7 \times 10=70$ cases, i.e., $1000,1004, \cdots$, 1980.

Therefore, there are $18+70=88$ four-digit numbers that meet the condition and have 1 as the thousandth digit.

Since $4+63+88=155$, the 155th number that meets the condition, in ascending order, is 1980.

The numbers that meet the condition below are 2000,2004, 2008, so the 158th number in this sequence is 2008.