Singapore Mathematical Olympiad (SMO)

Let the numbers be $X_1, X_2, \dots, X_7$ in ascending order. Let $[a, d]$ denote the arithmetic progression (AP) with initial term $a$, common difference $d$ and $a \le d$.

We first show that there is an AP $[a, d]$ that contains $X_i$, i.e. $X_i = a + k_i d$, $i = 1, \dots, 5$. For example, we can take $a = d = 1$. Choose such an AP with maximal $d$. Then not all $k_i$ can be even, or else we can use the AP $[a, 2d]$, contradicting the choice of $d$. Also, not all $k_i$ can be odd, or else we can use $[a+d, 2d]$.

We have two cases.

(i) At least 3 of $k_1, \dots, k_5$ are odd: use $[a+d, 2d]$.

(ii) At least 3 of $k_1, \dots, k_5$ are even: use $[a, 2d]$.

In either case, we have an AP $[a', d']$ that contains exactly 3 or 4 of $X_1, \dots, X_5$. Note that $[a', d']$ is a 'sub-AP' of $[a, d]$.

Next, we show that there is an AP that contains exactly 3 or 4 of $X_1, \dots, X_6$. If $[a', d']$ contains exactly 3 of $X_1, \dots, X_5$, or does not contain $X_6$, then it contains exactly 3 or 4 of $X_1, \dots, X_6$. We are now left with the case where $[a', d']$ contains exactly 5 of $X_1, \dots, X_6$. Then we can apply the above procedure to find a 'sub-AP' $[a'', d'']$ that contains exactly 3 or 4 of $X_1, \dots, X_6$.

The same procedure can be used again to find a 'sub-AP' of $[a'', d'']$ that contains exactly 3 or 4 of $X_1, \dots, X_7$.