Let the seven squares be $a_1^2, a_2^2, \ldots, a_7^2$ where $a_i$ are positive integers.

Consider the possible residues of $a^2$ modulo $20$.

Since $a$ can be any integer, $a^2$ modulo $20$ depends on $a$ modulo $20$.

Let us compute $a^2 \bmod 20$ for $a = 0, 1, \ldots, 19$:

Counting the number of distinct residues:
- $0$ (for $a \equiv 0, 10 \pmod{20}$)
- $1$ (for $a \equiv 1, 9, 11, 19 \pmod{20}$)
- $4$ (for $a \equiv 2, 8, 12, 18 \pmod{20}$)
- $5$ (for $a \equiv 5, 15 \pmod{20}$)
- $9$ (for $a \equiv 3, 7, 13, 17 \pmod{20}$)
- $16$ (for $a \equiv 4, 6, 14, 16 \pmod{20}$)

There are only $6$ possible residues for squares modulo $20$.

By the pigeonhole principle, among any $7$ squares, at least two must have the same residue modulo $20$.

Therefore, their difference is divisible by $20$.