Mongolian Mathematical Olympiad Round 3

Answer: $N = 56$.
First we prove $N = 56$ has the property. Suppose, on the contrary, that for a degree five polynomial $P(x)$ with integer coefficients, we have $|P(x)| \le \frac{1}{56}$ for all $0 \le x \le 1$.
Then the nonnegative integers $|P(0)|, |P(1)|, 32|P(1/2)|$ are strictly smaller than 1, thus are equal to 0. Let $Q(x) = x(1-x)(2x-1)$. By the Gauss lemma, there is a quadratic polynomial $D(x)$ with integer coefficients such that $P(x) = Q(x)D(x)$. Now consider
$$\alpha = \frac{1}{2} - \frac{1}{2\sqrt{5}} \quad \text{and} \quad \beta = \frac{1}{2} + \frac{1}{2\sqrt{5}}.$$
Clearly $0 < \alpha < \beta < 1$, $\alpha + \beta = 1$, $\alpha\beta = \frac{1}{5}$ and $Q(\alpha) = -\frac{1}{5\sqrt{5}}$, $Q(\beta) = \frac{1}{5\sqrt{5}}$.
Moreover, $25D(\alpha)D(\beta)$ is a polynomial expression of $5\alpha\beta$ and $\alpha + \beta$ with integer coefficients, thus is an integer. It follows
$$3125P(\alpha)P(\beta) = -25D(\alpha)D(\beta)$$
is an integer. On the other hand, $|3125P(\alpha)P(\beta)| \le \frac{3125}{3136} < 1$, hence $D(\alpha)D(\beta) = 0$. It follows $(5x^2 - 5x + 1) \mid D(x)$ and we may assume that $D(x) = 5x^2 - 5x + 1$. This leads to the contradiction:
$$P(0.9) = 0.9 \cdot 0.1 \cdot 0.8 \cdot 0.55 > 0.036 > 0.02 > \frac{1}{56}.$$
This proves for any degree five polynomial $P(x)$ with integer coefficients, there is $0 \le x \le 1$ such that $P(x) > \frac{1}{56}$.