Czech and Slovak Mathematical Olympiad

Notice that no negative number $m$ satisfies the problem evidently (absolute value is non-negative number).

Now we interpret the problem geometrically. A graph of some function $y = x^2 + a x + b$ lies in a horizontal strip between lines $y = +m$ and $y = -m$ and in the interval $(0, 2)$. Our aim is to find the closest strip which contains the graph of such quadratic function in the interval $(0, 2)$.

The function
$$f(x) = (x - 1)^2 - \frac{1}{2} = x^2 - 2x + \frac{1}{2},$$
seems to be a good candidate for such the closest strip. Such function has $a = -2$, $b = \frac{1}{2}$ and it satisfies (to be shown below) inequalities $-\frac{1}{2} \le f(x) \le \frac{1}{2}$.

Really, this inequalities are equivalent to the inequalities $0 \le (x-1)^2 \le 1$, which are evidently fulfilled for $x \in (0, 2)$. Quadratic function $f(x) = x^2 - 2x + \frac{1}{2}$ thus satisfies the conditions of the problem for $m = \frac{1}{2}$.

In the second part of the solution we will show that there is no quadratic function satisfying the problem for any $m < \frac{1}{2}$.

The crucial fact will be that at least one from differences $f(0) - f(1)$ and $f(2) - f(1)$ is greater or equal to $1$ for an arbitrary function $f(x) = x^2 + a x + b$. This fact will imply that width of the closest strip will be greater or equal to $1$. This will exclude the values $m < \frac{1}{2}$. For $f(0) - f(1) \ge 1$ we obtain the desired estimate $2m \ge 1$ easily from the well-known triangle inequality $|a - b| \le |a| + |b|$:
$$1 \le |f(0) - f(1)| \le |f(0)| + |f(1)| \le 2m.$$
Similarly we estimate for $f(2) - f(1) \ge 1$.

Now it remains to verify at least one from inequalities $f(0) - f(1) \ge 1$ and $f(2) - f(1) \ge 1$ for arbitrary $f(x) = x^2 + a x + b$. The values
$$f(0) = b, \quad f(1) = 1 + a + b, \quad f(2) = 4 + 2a + b,$$
yields
$$f(0) - f(1) = -1 - a \ge 1 \Leftrightarrow a \le -2,$$
$$f(2) - f(1) = 3 + a \ge 1 \Leftrightarrow a \ge -2.$$
So at least one from inequalities $f(0) - f(1) \ge 1$ and $f(2) - f(1) \ge 1$ is true (regardless of the choice $a, b$).

Conclusion. The desired minimal value of $m$ is $\frac{1}{2}$.