Let us analyze the game played by Susana and Brenda, where they write polynomials on the board.

### Rules Recap
1. Initially, Susana selects a positive integer $n_0$ and writes the polynomial $P_0(x) = n_0$.
2. On her first turn, Brenda must choose a positive integer $n_1 \neq n_0$ and write either:
$$P_1(x) = n_1x + P_0(x) \quad \text{or} \quad P_1(x) = n_1x - P_0(x).$$
3. For turn $k$, the player chooses a positive integer $n_k \neq n_0, n_1, \ldots, n_{k-1}$ and writes either:
$$P_k(x) = n_kx^k + P_{k-1}(x) \quad \text{or} \quad P_k(x) = n_kx^k - P_{k-1}(x).$$
4. The first player to write a polynomial with an integer root wins.

### Winning Strategy for Susana

1. Turn 0:
- Susana writes $P_0(x) = 1$.

2. Turn 1:
- Brenda, not allowed to use $n_1 = 1$, writes either:
$$P_1(x) = n_1x + 1 \quad \text{or} \quad P_1(x) = n_1x - 1.$$
- Neither of these polynomials have integer roots since for $P_1(x) = n_1x + 1$ or $P_1(x) = n_1x - 1$, solving for integer roots would require $x = -\frac{1}{n_1}$ or $x = \frac{1}{n_1}$, both of which are not integers as $n_1 \neq \pm 1$.

3. Turn 2:
- Depending on Brenda's choice in Turn 1:
- If Brenda wrote $P_1(x) = n_1x - 1$, Susana writes:
$$P_2(x) = (n_1 + 1)x^2 + n_1x - 1.$$
Substituting $x = -1$:
$$P_2(-1) = (n_1 + 1)(-1)^2 + n_1(-1) - 1 = (n_1 + 1) - n_1 - 1 = 0,$$
Thus, $-1$ is a root, and Susana wins.
- If Brenda wrote $P_1(x) = n_1x + 1$, Susana writes:
$$P_2(x) = (n_1 + 1)x^2 - n_1x - 1.$$
Substituting $x = 1$:
$$P_2(1) = (n_1 + 1)(1)^2 - n_1(1) - 1 = (n_1 + 1) - n_1 - 1 = 0,$$
Thus, $1$ is a root, and Susana wins.

Based on this strategy, Susana ensures that no matter what polynomial Brenda writes on Turn 1, Susana can always write a polynomial on Turn 2 that has an integer root. Therefore, the solution allows Susana to win the game by setting up the situation in such a way that Brenda's choices do not affect the outcome by Turn 2.

The strategy can be concluded with the realization that Susana has a winning path by carefully managing her choices:

$$\boxed{\text{Susana wins by writing } P_2(x) \text{ that has } +1 \text{ or } -1 \text{ as a root.}}$$