, 2012

Let $P_k$ be the product of all the numbers on the chalkboard after the $k$$\text{th}$$step. If in the$k$\text{th}$ step we choose the numbers $a > b$ such that $b$ divides $a$, then
$$P_k = \frac{P_{k-1}}{ab} \cdot \frac{a}{b} = \frac{P_{k-1}}{b^2}.$$
Thus, after each step, the product of all the numbers on the chalkboard is divisible by a square of a natural number. That is, the parity of the exponents in the prime factorization of the product of all the numbers on the chalkboard does not change. Because at the beginning the product is
$$P_0 = 1 \cdot 2 \cdots 33 = 2^{31} \cdot 3^{15} \cdot 5^7 \cdot 7^4 \cdot 11^3 \cdot 13^2 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31,$$
the product of all the numbers that stay on the chalkboard is a multiple of the number $2 \cdot 3 \cdot 5 \cdot 11 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Since the prime numbers $17$, $19$, $23$, $29$ and $31$ do not divide any other number on the chalkboard, they stay on the chalkboard till the end. The product of all the other numbers on the chalkboard at the end is at least $2 \cdot 3 \cdot 5 \cdot 11 = 330 > 33$. Thus, at the end, there must be two additional prime numbers that are greater than $16$ on the chalkboard. Altogether there must be at least $7$ such prime numbers.

We can get $7$ numbers in the described process as follows. First successively erase the following pairs:
$(33, 11)$, $(25, 5)$, $(27, 9)$, $(28, 14)$, $(26, 13)$, $(21, 7)$, $(30, 10)$, $(20, 5)$, $(32, 16)$, $(24, 12)$, $(18, 6)$, $(8, 4)$, $(3, 3)$, $(3, 3)$, $(3, 3)$, $(4, 2)$, $(2, 2)$, $(2, 2)$, $(2, 2)$. Then erase all the ones to get the numbers $15$, $17$, $19$, $22$, $23$, $29$, $31$.