Let $N$ be the product of all the odd integers lying in between $1$ and $100$. It is clear that the last $3$ digits of $N$ equals the remainder obtained when $N$ is divided by $1000$. Note that $1000 = 10^3 = 2^3 \times 5^3 = 8 \times 125$.

From the identity $(8x + a)(8y + b) = 8(axy + a + b) + ab$, valid for any choice of integers $x, y, a, b$, it follows that for any integer $P$ which can be written as a product $P_1P_2$ of integers $P_1$ and $P_2$, the remainder obtained when $P$ is divided by $8$ is the same as the product of the remainders obtained when $P_1$ and $P_2$ are divided by $8$, respectively. Therefore, the remainder obtained by dividing $N$ by $8$ must be the product of the remainders obtained when each odd integer lying in between $1$ and $100$ are divided by $8$.

When odd numbers $1, 3, 5, 7, 9, 11, \dots$ are divided by $8$ the remainders $1, 3, -3, -1, 1, \dots$ are obtained, and one can see easily that the block of $4$ numbers $1, 3, -3, -1$, repeats $12$ times and followed by $1, 3$ (as $97 = 8 \times 12 + 1$ and $99 = 8 \times 13 + 3$). Therefore, we can conclude that the remainder obtained when $N$ is divided by $8$ equals the remainder obtained by dividing $3^{25}$ by $8$, and this number equals $3$, since $3^{25} = 9^{12} \times 3$, and the remainder obtained by dividing $9$ by $8$ is $1$.

Furthermore, since $1000$ is divisible by $8$, the remainder obtained when the last $3$-digits of $N$ is divided by $8$ must also be $3$. From these considerations we conclude that the solution to the problem is given by the number between $0$ and $999$ which is a multiple of $125$ and gives a remainder $3$ when divided by $8$. The number $875$ is the only number satisfying these requirements and is the desired answer.