Problem 803

Initially observe that, from 1 to 77, Joana erased 11 multiples of 7 and 7 multiples of 11. Since 77 is a multiple of 7 and 11, she erased $11+7-1=17$ numbers, leaving $77-17=60$ numbers. Now, grouping the first 10000 numbers into groups of 77 consecutive numbers, this reasoning applies to each of the lines below, that is, in each line 60 numbers remained.

$$\begin{array}{ccccc} \text { 1st line: } & 1, & 2, & \ldots, & 77 \\ 2^{\text {nd }} \text { line: } & 78, & 79, & \ldots, & 154 \\ 3^{\text {rd }} \text { line: } & 155, & 158, & \ldots, & 231 \end{array}$$

Since $2008=33 \times 60+28$, we know that among the first $33 \times 77=2541$ numbers, $33 \times 60=1980$ numbers remained without being erased.

$$\text { 33rd line: } \quad \ldots, \quad \ldots, \quad \ldots, \quad 2541$$

We still need to count 28 numbers. Let's then examine the 34th line, which starts with 2542. Since the erased numbers are in columns $7,11,14,21,22,28,33,35$, etc., and up to the $35^{\text {th }}$ column, eight numbers have been erased, there remain $35-8=27$ numbers in the $34^{\text {th }}$ line. Therefore, after erasing the multiples of 7 and 11 in this line, the 28th number is 2577. Thus, the number in the 2008th position is 2577.