A six-digit almost palindrome can be written as $abcbad$, where $a, d \neq 0$. This number is a multiple of $9$ if and only if the sum $a + b + c + b + a + d$ is a multiple of $9$. Notice that if the values of $a, b$ and $c$ are fixed, then the remainder of $d$ when divided by $9$ is determined. We know that $d$ can be any digit from $1$ to $9$. Since those digits have different remainders when divided by $9$, and every possible remainder is attained by one of those digits, then no matter how we choose $a, b$ and $c$, there is always exactly one possible value for $d$ that makes $abcbad$ a multiple of $9$. Finally, since $a$ can be any digit between $1$ and $9$ and $b$ and $c$ can be any digit between $0$ and $9$, there are exactly $9 \times 10 \times 10 = 900$ six-digit almost palindrome numbers divisible by $9$.