Since $250 = 2 \cdot 5^3$, Andrew has to cross out from the product $1 \cdot 2 \cdots 100$ all numbers which are divisible by $5$, except for two numbers that are not divisible by $25$ (for example, we can leave $5$ and $10$). The resulting product satisfies the condition because it is not divisible by $5^3 = 125$. Meanwhile, Andrew crossed out $18$ numbers, because among the factors $1, 2, \ldots, 100$ exactly $20$ are divisible by $5$.

Let us assume that it is possible to cross out no more than $17$ numbers. Thus, among $20$ numbers that are divisible by $5$, Andrew left at least three, then the product is divisible by $5^3$. In order for the product not to be divisible by $250$, all factors must be odd. However, since $83$ numbers are left and only $50$ initial numbers were odd, at least one of the factors is even. Contradiction completes the proof.