Japanese Mathematical Olympiad - The First Round

It is clear that the smallest value exists. Let $B$ be a book with $n_i$ pages for the $i$-th chapter which attains the smallest value.

We first prove that none of $n_1, n_2, \dots, n_{10}$ are equivalent to $3$ or $5$ modulo $6$. Assume that $n_1 \equiv 3, 5 \pmod{6}$. Checking the parity, we can assume without loss of generality that $n_2$ is odd. Consider book $B'$ whose chapter one has 4 pages, chapter two has $n_1 + n_2 - 4$ pages, and the remaining chapters have the same number of pages with corresponding chapters of $B$. Then easy computation shows that Akinori can read $B'$ faster than $B$, and that Tomohiro can't read $B'$ faster than $B$. Then the difference of the days needed to read $B'$ is smaller than that of $B$, which contradicts minimality of $B$.

Now let $n_i = 6m_i + k_i$ ($m_i$ nonnegative, $k_i$ nonnegative $< 6$ and $k_i \ne 3, 5$). Then it can be easily checked that Tomohiro reads $i$-th chapter $m_i$ days faster than Akinori. Since
$$\sum_{i=1}^{10} m_i = \frac{1}{6} \left( \sum_{i=1}^{10} n_i - \sum_{i=1}^{10} k_i \right) = \frac{1}{6} \left( 120 - \sum_{i=1}^{10} k_i \right)$$
and $k_i \le 4$, it follows that $\sum m_i \ge 20 - \frac{40}{6}$, hence $\sum m_i \ge 14$. This value is in fact attained if, for example, $n_1 = n_2 = \dots = n_9 = 10$ and $n_{10} = 30$. Therefore, the smallest value is 14.