Romanian Mathematical Olympiad

a) If $f([a, b]) \subseteq [0, \infty)$ or $f([a, b]) \subseteq (-\infty, 0]$, then $m = |f(\frac{a+b}{2})| > 0$, and on one of the intervals $(a, \frac{a+b}{2})$ or $(\frac{a+b}{2}, b)$ the inequality $|f(x)| > m$ holds for any $x$ in that interval. Then
$$\begin{aligned} 0 &= \left| \int_a^b f(x) \, dx \right| = \int_a^b |f(x)| \, dx = \int_a^{\frac{a+b}{2}} |f(x)| \, dx + \int_{\frac{a+b}{2}}^b |f(x)| \, dx \\ &\ge m \cdot \frac{b-a}{2} > 0, \end{aligned}$$
which is impossible. It follows that $f(a) \cdot f(b) < 0$.

b) We will show that the only sequences of real numbers $(a_n)_{n \ge 1}$ which satisfy the condition in the statement are the constant sequences and the sequences assuming exactly two distinct real values, which become stationary beginning with a certain rank.
Obviously, if $(a_n)_{n \ge 1}$ is a constant sequence, with $a_n = a$, $a \in \mathbb{R}$, then the sequence is convergent, with $\lim_{n \to \infty} a_n = a$ and $\int_{a_k}^{a_{k+1}} f(x) dx = 0$, for any $k \ge 1$, and any function $f : \mathbb{R} \to \mathbb{R}$. If $\{a_n | n \ge 1\} = \{a, b\}$ and there is a rank $n_0 \in \mathbb{N}^*$ with $a_n = a$, for any $n \ge n_0$, then $(a_n)_{n \ge 1}$ is convergent, with $\lim_{n \to \infty} a_n = a$, and there is the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x - a - b$, which is strict monotone and satisfies the equalities $\int_{a_k}^{a_{k+1}} f(x) dx = 0$, for any $k \ge 1$.

Let $(a_n)_{n \ge 1}$ be a convergent sequence of real numbers for which there is a strict monotone function $f : \mathbb{R} \to \mathbb{R}$ such that
$$\int_{a_1}^{a_2} f(x) \, dx = \int_{a_2}^{a_3} f(x) \, dx = \dots = \int_{a_n}^{a_{n+1}} f(x) \, dx = \dots$$

Let $I = \int_{a_k}^{a_{k+1}} f(x) dx$, for every $k \ge 1$, and $a = \lim_{n \to \infty} a_n$. For a fixed $r > 0$ there is a rank $n_r \ge 1$, such that $a_n \in (a - r, a + r)$, for any $n \ge n_r$. Since $f$ is strict monotone, if $M = \max(|f(a - r)|, |f(a + r)|)$, then $|f(x)| < M$, for every $x \in (a - r, a + r)$. Then, for any $p \in \mathbb{N}^*$ we have:
$$\begin{aligned} p \cdot |I| &= |p \cdot I| = \left| \sum_{k=1}^{p} \int_{a_{n_r+k-1}}^{a_{n_r+k}} f(x) \, dx \right| = \left| \int_{a_{n_r}}^{a_{n_r+p}} f(x) \, dx \right| \\ &\le \left| \int_{a_{n_r}}^{a_{n_r+p}} |f(x)| \, dx \right| < \int_{a-r}^{a+r} |f(x)| \, dx \le 2r \cdot M. \end{aligned}$$
It follows that $0 \le |I| < \frac{2Mr}{p}$, for any $p \in \mathbb{N}^*$, so that $I = 0$.
We show now that $\text{card}(\{a_n | n \in \mathbb{N}^*\}) \le 2$. Assuming the opposite, there exist $i, j, k \in \mathbb{N}^*$, with $i < j < k$ such that $a_i \ne a_j \ne a_k \ne a_i$. Then
$$\int_{a_i}^{a_j} f(x) \, dx = \sum_{l=i}^{j-1} \int_{a_l}^{a_{l+1}} f(x) \, dx = (j-i) \cdot I = 0$$
and, also, $\int_{a_j}^{a_k} f(x) \, dx = (k-j) \cdot I = 0$, respectively $\int_{a_i}^{a_k} f(x) \, dx = 0$.
The function $f$ being strict monotone, it follows that $f(a_i) \cdot f(a_j) < 0$, $f(a_i) \cdot f(a_k) < 0$, and $f(a_j) \cdot f(a_k) < 0$. But then
$$(f(a_i) \cdot f(a_j) \cdot f(a_k))^2 = (f(a_i) \cdot f(a_j)) \cdot (f(a_i) \cdot f(a_k)) \cdot (f(a_j) \cdot f(a_k)) < 0,$$
which is absurd.