Suppose that $S_1, S_2, \dots, S_{18}$ leave the same remainder when divided by 5. Since $S_1 = a_1 + (a_2 + a_3 + a_4 + a_5)$ and $S_2 = (a_2 + a_3 + a_4 + a_5) + a_6$, if $S_1$ and $S_2$ leave the same remainder when divided by 5, then $a_1$ and $a_6$ leave the same remainder when divided by 5. In the same way:

1. $a_1, a_6, a_{11}, a_{16}$ and $a_{21}$ leave the same remainder, $x$, when divided by 5.
2. $a_2, a_7, a_{12}, a_{17}$ leave the same remainder, $a$, when divided by 5.
3. $a_3, a_8, a_{13}, a_{18}$ leave the same remainder, $b$, when divided by 5.
4. $a_4, a_9, a_{14}, a_{19}$ leave the same remainder, $c$, when divided by 5.
5. $a_5, a_{10}, a_{15}, a_{20}$ leave the same remainder, $d$, when divided by 5.

Since $\{a_1, a_2, \dots, a_{21}\} = \{1, 2, \dots, 21\}$, there are 5 remainders equal to 1 and 4 remainders equal to each of 2, 3, 4 or 0.
It follows $x = 1$ and $\{a, b, c, d\} = \{0, 2, 3, 4\}$.

Now $S_1 = \mathcal{M}5 + 1 + 0 + 2 + 3 + 4 = \mathcal{M}5$ and $S_{18} = \mathcal{M}5 + b + \mathcal{M}5 + c + \mathcal{M}5 + d + \mathcal{M}5 + 1 + \mathcal{M}5 + 1 = \mathcal{M}5 + 2 + (a + b + c + d) - a = \mathcal{M}5 + 11 - a$.

Since $S_1$ and $S_{18}$ leave the same remainder when divided by 5, it follows that $11 - a = \mathcal{M}5$, whence $a = 1$ – contradiction.