The number 119 has the following property:

- division by 2 leaves a remainder of 1;
- division by 3 leaves a remainder of 2;
- division by 4 leaves a remainder of 3;
- division by 5 leaves a remainder of 4;
- division by 6 leaves a remainder of 5.

How many positive integers less than 2007 satisfy this property?
Initially note that if $N$ divided by $d$ leaves a remainder $r$, then adding a multiple of $d$ to $N$ does not change the remainder, that is:

$$
\frac{(N+\text { multiple of } d)}{d} \text { also leaves a remainder of } r
$$

For example: 38 divided by 3 leaves a remainder of 2, so the remainder of the division of $(38+5 \times 3)$ is also 2.

Thus, if we add to 119 a number that is a multiple of 2, 3, 4, 5, and 6 simultaneously, this number leaves the same remainders as 119 when divided by $2, 3, 4, 5,$ and 6. The least common multiple of $2, 3, 4, 5,$ and 6 is 60, so every number of the form

$$
119+(\text { multiple of 60) }
$$

satisfies the five conditions stated in the problem.

From the division of 2007 by 60, we have:

$$
2007=33 \times 60+27=32 \times 60+87=31 \times 60+147
$$

Since 119 is between 87 and 147, we have that the numbers

$$
59,119,179, \ldots, 31 \times 60+119
$$

satisfy the same property as 119. Therefore, we have 33 possible numbers.