Let's denote the last three digits of a positive integer as $x$, and the number formed by the digits preceding the last three as $y$.

Assume $x - y$ is a multiple of 7, meaning $x - y$ is divisible by 7.

Then, the integer can be expressed as: $1000y + x = 1001y + (x - y)$,

Since $1001y = 143y \times 7$,

Both $1001y$ and $(x - y)$ are multiples of 7,

Therefore, $1000y + x$ is a multiple of 7,

This implies that the positive integer is divisible by 7.

Assume $x - y$ is a multiple of 11, meaning $x - y$ is divisible by 11.

Then, the integer can be expressed as: $1000y + x = 1001y + (x - y)$,

Since $1001y = 91y \times 11$,

Both $1001y$ and $(x - y)$ are multiples of 11,

Therefore, $1000y + x$ is a multiple of 11,

This implies that the positive integer is divisible by 11.

Thus, we have proved that if the difference between a positive integer's last three digits and the number formed by the digits preceding the last three is divisible by 7 (or 11), then the positive integer is divisible by 7 (or 11). The final answer is encapsulated as $\boxed{\text{Proved}}$.