Each of the positive integers 2013 and 3210 has the following three properties:

(i) it is an integer between 1000 and 10000 ,

(ii) its four digits are consecutive integers, and

(iii) it is divisible by 3 .

In total, how many positive integers have these three properties?
We recall that a positive integer is divisible by 3 whenever the sum of its digits is divisible by 3 . Since the sum of the digits does not depend on the order of the digits, then rearranging the digits of a positive integer that is divisible by 3 produces another positive integer that is divisible by 3 .

Note that 10000 is not divisible by 3. Every other positive integer between 1000 and 10000 is a four-digit integer.

Consider a four-digit positive integer whose four digits are consecutive integers. We can rearrange the digits of this integer in decreasing order to obtain one of the positive integers 3210 , $4321,5432,6543,7654,8765$, and 9876 .

The sums of the digits of these integers are $6,10,14,18,22,26$, and 30 , respectively.

Of these, 6,18 and 30 are the only sums that are divisible by 3 , so 3210,6543 and 9876 are the only ones divisible by 3 .

Since rearranging the digits does not affect whether an integer is divisible by 3, then a four-digit integer satisfies the given conditions if its digits are rearrangements of 3210 or 6543 or 9876 .

There are 24 four-digit integers whose digits are rearrangements of 6543 . (There are 4 possibilities for the thousands digit, then 3 possibilities for the hundreds digit, then 2 possibilities for the tens digit, and 1 possibility for the units digit, and so $4 \times 3 \times 2 \times 1=24$ integers that use these digits.)

Similarly, there are 24 four-digit integers whose digits are rearrangements of 9876 .

Finally, there are 18 four-digit integers whose digits are rearrangements of 3210. (There are 3 possibilities for the thousands digits (because 0 cannot be the thousands digit), then 3 possibilities for the hundreds digit, 2 for the tens digit, and 1 for the units digit, and so $3 \times 3 \times 2 \times 1=18$ integers that use these digits.)

In total, there are $24+24+18=66$ positive integers with these three properties.

ANSWER: 66

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