Italian Mathematical Olympiad

The answer is 1617. If $a, b, c, d$ is an arithmetic progression of four terms, then the common difference $r$ is also an integer $\geq 1$ and we have $d = a + 3r$. On the other hand, the first term and the last term of an arithmetic progression of four terms completely determine the other terms as well. The number of arithmetic progressions sought is therefore in bijective correspondence with the pairs of numbers of the form $\{a, a+3r\}$ contained in the set $\{1, \ldots, 100\}$. Observing that $a$ and $a+3r$ give the same remainder upon division by 3, we can distinguish 3 cases, according to whether this remainder is 1, 2, or 0:

- $\{a, a+3r\} \subset X_{1} = \{1, 4, 7, \ldots, 100\}$. The set $X_{1}$ consists of 34 elements, so there are $\binom{34}{2} = 561$ pairs of this type.
- $\{a, a+3r\} \subset X_{2} = \{2, 5, 8, \ldots, 98\}$. The set $X_{2}$ consists of 33 elements, so there are $\binom{33}{2} = 528$ pairs of this type.
- $\{a, a+3r\} \subset X_{3} = \{3, 6, 9, \ldots, 99\}$. The set $X_{3}$ consists of 33 elements, so there are $\binom{33}{2} = 528$ pairs of this type.

In total, $561 + 528 + 528 = 1617$ arithmetic progressions.

Second solution.
Denoting by $r \geq 1$ the common difference of the arithmetic progression, we have $1 \leq a$ and $a + 3r \leq 100$, so $3r \leq 100 - 1 = 99$ and $r \leq 33$. If $a, a + r, a + 2r, a + 3r$ is an arithmetic progression with common difference $r$, we have $a \geq 1$ and $a + 3r \leq 100$, that is $a \leq 100 - 3r$; therefore there are exactly $100 - 3r$ arithmetic progressions with common difference $r$. The total number of arithmetic progressions is therefore
$$\sum_{r=1}^{33} (100 - 3r) = 100 \cdot 33 - 3 \sum_{r=1}^{33} r = 3300 - 3 \frac{33 \cdot 34}{2} = 3300 - 1683 = 1617.$$