Since $N$ is between 1000000 and 10000000, inclusive, then $25 \times N$ is between 25000000 and 250000000, inclusive, and so $25 \times N$ has 8 digits or it has 9 digits. We consider the value of $25 \times N$ as having 9 digits, with the possibility that the first digit could be 0. Since $25 \times N$ is a multiple of 25, its final two digits must be $00,25,50$, or 75. For a fixed set of leftmost three digits, $xyz$, the multiple of 25 that has the largest sum of digits must be $xyz999975$ since the next four digits are as large as possible (all 9s) and the rightmost two digits have the largest possible sum among the possible endings for multiples of 25. So to answer the question, we need to find the integer of the form $xyz999975$ which is between 25000000 and 250000000 and has the maximum possible sum $x+y+z$. We know that the maximum possible value of $x$ is 2, the maximum possible value of $y$ is 9, and the maximum possible value of $z$ is 9. This means that $x+y+z \leq 2+9+9=20$. We cannot have 299999975 since it is not in the given range. However, we could have $x+y+z=19$ if $x=1$ and $y=9$ and $z=9$. Therefore, the integer 199999975 is the multiple of 25 in the given range whose sum of digits is as large as possible. This sum is $1+6 \times 9+7+5=67$. We note that $199999975=25 \times 7999999$ so it is a multiple of 25. Note that $N=7999999$ is between 1000000 and 10000000.