Problem 1985

Let's look at the sum of the triple. This does not change after each step since $(y+z-x)+(z+x-y)+(x+y-z) = x+y+z$. Moreover, the values of the new triple do not depend on the order of the numbers in the original triple, since the terms are symmetrical in all three variables. Hence we may assume $a < b < c$ and we can change the order of the numbers in the triple.

Solution 1: After the first step we get the triple $(a + b - c, a + c - b, b + c - a)$ and we have
$$a + b - c < a + c - b < b + c - a,$$
after the second step we get the triple $(3a - b - c, 3b - a - c, 3c - a - b)$ and we have
$$3a - b - c < 3b - a - c < 3c - a - b.$$

Let's look at the differences between individual numbers of the triple. By induction we check that after $n$ steps we have the following:
* The difference between the smallest and the largest number in the triple is equal to $2^n(c-a)$.
* The difference between the smallest and the middle number in the triple is equal to $2^n(b-a)$ if $n$ is even and $2^n(c-b)$ if $n$ is odd.
* The difference between the middle and the largest number in the triple is equal to $2^n(c-b)$ if $n$ is even and $2^n(b-a)$ if $n$ is odd.

After $n$ steps we have the triple of the form $(t, t + 2^n(b-a), t + 2^n(c-a))$ if $n$ is even and $(t, t + 2^n(c-b), t + 2^n(c-a))$ if $n$ is odd. Here $t$ is some integer which can be calculated from the condition on the sum. We have
$$\begin{aligned} 2015 &= t + (t + 2^n(b-a)) + (t + 2^n(c-a)) = 3t + 2^n(b+c-2a) \\ &\Rightarrow t = \frac{2015 - 2^n(b+c-2a)}{3}, \text{ if } n \text{ even, and} \end{aligned}$$
$$\begin{aligned} 2015 &= t + (t + 2^n(c-b)) + (t + 2^n(c-a)) = 3t + 2^n(2c-a-b) \\ &\Rightarrow t = \frac{2015 - 2^n(2c-a-b)}{3}, \text{ if } n \text{ odd.} \end{aligned}$$
If any of the numbers in the triple will be negative, it will surely be $t$. Since the sum of three consecutive numbers cannot equal $2015$, the difference between the largest and the smallest number in the initial triple is at least $3$. This gives $c-a \ge 3$. Thus we can estimate $b+c-2a = (c-a)+(b-a) \ge 4$ and $2c-a-b = (c-a)+(c-b) \ge 4$. It follows
$$t \le \frac{2015 - 4 \cdot 2^n}{3}.$$

With the initial triple $(670, 672, 673)$ we get $t = \frac{2015-2^8(673+672-2 \cdot 670)}{3} = 245$ after $8$ steps. Therefore this triple does not yield a negative number after $8$ steps.
So we have to make at least $9$ steps to get a triple with at least one negative number.

Solution 2: Since $a+b+c = 2015$, after the first step we get the triple $(b+c-a, a+c-b, a+b-c) = (2015-2a, 2015-2b, 2015-2c)$. Thus in each step the transformation acts component-wise by the rule $x \mapsto 2015-2x$.
We can check by induction that if we apply this transformation $n$ times to the number $x$ we get
$$(-2)^n x + \frac{1-(-2)^n}{3} \cdot 2015.$$

Let's determine when this number is non-negative. For even $n$ we must have
$$\begin{aligned} 2^n x + \frac{1-2^n}{3} \cdot 2015 &\ge 0 \\ \Leftrightarrow \frac{1}{\frac{2015}{3}-x} \cdot \frac{2015}{3} &\ge 2^n. \end{aligned}$$

For odd $n$ we must have
$$\begin{aligned} -2^n x + \frac{1+2^n}{3} \cdot 2015 &\ge 0 \\ \Leftrightarrow \frac{1}{x - \frac{2015}{3}} \cdot \frac{2015}{3} &\ge 2^n. \end{aligned}$$

Since $2015$ is not a sum of three consecutive positive integers, we must have $a \le 670$ and $c \ge 673$. Inserting this in the above inequalities we get
$$2^n \le \frac{1}{\frac{2015}{3} - a} \cdot \frac{2015}{3} \le \frac{1}{\frac{2015}{3} - 670} \cdot \frac{2015}{3} = 403 \Rightarrow n \le 8,$$
$$2^n \le \frac{1}{c - \frac{2015}{3}} \cdot \frac{2015}{3} \le \frac{1}{673 - \frac{2015}{3}} \cdot \frac{2015}{3} = \frac{2015}{4} \Rightarrow n \le 8.$$

So we have to make at least $9$ steps to get a triple with at least one negative number.