Gonna Need A Big Chalkboard

Relevant wiki: Lowest Common Multiple

Observe that $\text{lcm} \{a, b, c\} = \text{lcm} \{ \text{lcm} \{a, b\}, c\}$ ; in other words, the final result doesn't depend on the order we take the LCMs, because they commute. Thus the final LCM is wholly determined by the numbers we pick initially.

There are 15 multiples of 128 below 2017. If we don't take any of them, then the LCM cannot be a multiple of 128, since the highest power of 2 that divides the LCM would be 64. (Remember that in order to have the LCM divisible by $2^7 = 128$ , one of the numbers must be a multiple of $2^7 = 128$ . This would be different if the LCM had to be divisible by a non-prime power, such as 6; in that case, we need one number divisible by 2, and another (not necessarily different) divisible by 3; we don't necessarily need a number that is divisible by both 2 and 3 simultaneously. Note also that this would be different had the LCM been "product".)

Since we can skip 15 numbers and have LCM that is not a multiple of 128, we need to take at least one of these numbers. That gives $2017 - 15 + 1 = \boxed{2003}$ numbers that we need to take. This is clearly enough; taking 2003 numbers will require us to take some multiple of 128, because otherwise there are only 2002 numbers available.

Relevant wiki: Lowest Common Multiple

If $x_1, x_2, x_3, \ldots , x_m$ are the $m$ randomly chosen positive integers, then this process merely calculates $lcm(x_1, x_2, x_3, \ldots , x_m)$ . Set $L=lcm(x_1, x_2, x_3, \ldots , x_m)$ .

Then we are looking for the smallest $m$ such that $128$ necessarily divides $L$ . Since the $x_i$ are randomly chosen, $m$ must be large enough to exhaust all non-multiples of $128$ . Since there are $\lfloor\frac{2017}{128} \rfloor=15$ multiples of $128$ , we know there are $2002$ non-multiples. Therefore, the smallest $m$ is $2003$ .

Relevant wiki: Lowest Common Multiple

Let the final number be $F$ and the $n^{th}$ chosen number be written in the form $a_{n} \times 2^{b_{n}}$ where $a_{n},b_{n}$ are integers and $b_{n}$ is as large as possible. This implies that $a_{n}$ is odd, since if it were even, $b_{n}$ is not maximized.

Now, taking the least common multiple of two of these numbers, call them $x=p \times 2^{j}$ and $y=q \times 2^{k}$ , will give us $LCM(x,y)=LCM(p,q) \times 2^{\max(j,k)}$ . Since $p,q$ are both odd, their least common multiple will be odd and thus has no impact on divisibility by $128=2^{7}$ .

Hence, for $128|F$ , we must have $\max(b_{1},b_{2},b_{3},...,b_{m}) \geq 7$ . In other words, $128|F$ if and only if any of the chosen numbers are a multiple of $128$ .

There are $15$ positive multiples of $128$ less than $2017$ (we know this because $128|(2048-128)=15 \times 2^{7}$ ) and so we can choose all but $15$ of the possible numbers without forcing the condition to be met. Hence, we can choose $2002$ but not $\boxed{2003}$ .

Bonus: $m = 2017 - \lfloor\frac{2017}{n}\rfloor$ + 1