jmc

We can use the distributive property of multiplication to multiply a three-digit palindrome $aba$ (where $a$ and $b$ are digits) with 101: $$101\cdot aba=(100+1)\cdot aba=aba00+aba=ab(2a)ba.$$Here, the digits of the product are $a$, $b$, $2a$, $b$, and $a$, unless carrying occurs. In fact, this product is a palindrome unless carrying occurs, and that could only happen when $2a\ge 10$. Since we want the smallest such palindrome in which carrying occurs, we want the smallest possible value of $a$ such that $2a\ge 10$ and the smallest possible value of $b$. This gives us $\boxed{505}$ as our answer and we see that $101\cdot 505=51005$ is not a palindrome.