Let $n$ be the desired number, and let $m$ be the number obtained by deleting the first digit. Since the last digit of the product $m \cdot 25$ is not 0, it can only be 5. Since $5 \cdot 25$ has three digits, it cannot be the number $n$ itself,
so the number $m$ must have more digits. Consider a digit $d$ in both $n$ and $m$. Then in the product $d$ is obtained from multiplying all lower-order digits by 25 and adding the last digit of the number $d \cdot 5$ to the corresponding digit in the product. If $d$ is even then the corresponding digit in the product of lower order digits by 25 is $d$, i.e. an even number, while in the case of odd $d$, it is $d - 5$ or $d + 5$, i.e. again an even number.
Since the tens digit of $5 \cdot 25$ is 2, the tens digit of $n$ and $m$ must be either 2 or 7. If it is 2 and $m$ does not have more digits, then $n = 25 \cdot 25 = 625$ which satisfies the conditions of the problem. If there are more digits in the number $m$, then the hundreds digit of the numbers $m$ and $n$ must be 6 or 1, because the hundreds digit of the intermediate result of multiplying lower-order numbers is 6. Now $n$ cannot be $625 \cdot 25$ because the product has 5 digits. Since the thousands digit 5 of this product is odd, it is not possible to add more digits to the number $m$. If the hundreds digit is 1 then we get the possibility $m = 125$ and $n = 3125$, which satisfies the conditions of the problem, and since the thousands digit in this number is odd, it is not possible to add more digits to the number $m$ here either.
If the tens digit of $m$ and $n$ is 7, then $n$ cannot be $75 \cdot 25$ because the product has 4 digits. Since the hundreds digit of $75 \cdot 25$ is 8, the hundreds digit of $m$ can be 8 or 3. Again $n$ cannot be $875 \cdot 25$ because the product has 5 digits, and since the thousands digit 1 of the product is odd, it is not possible to add more digits to the number $m$. If the hundreds digit of $m$ is 3 we get $m = 375$ and $n = 9375$, which satisfies the conditions of the problem. Since the thousands digit of the product is odd, it is not possible to add more digits to the number $m$ here either.
Thus only 625, 3125 and 9375 satisfy the conditions of the problem.

Let $n$ be the desired number. Let the first digit of this number be $a$ and the number formed from the remaining digits be $m$, with the number $m$ having $k$ digits. Then $n = 10^k a + m = 25m$, implying $10^k a = 24m$. Thus, the number $10^k a = 2^k 5^k a$ is divisible by the number 24. Consequently, the number $2^k 5^k a$ must be divisible by the numbers $2^3$ and $3$. Since the number $2^k 5^k$ is not divisible by 3, the number $a$ must be divisible by 3, which leaves the possibilities $a = 3, a = 6$ and $a = 9$.
* If $a = 3$, then $2^k 5^k$ must be divisible by $2^3$. Therefore $k \ge 3$ and we get $m = \frac{10^k a}{2^4} = \frac{3000 \cdot 10^{k-3}}{2^4} = 125 \cdot 10^{k-3}$. Since last digit of $m$ is not zero, we have $m = 125$, whence $n = 3125$.
* If $a = 6$, then $2^k 5^k$ must be divisible by $2^2$. Therefore $k \ge 2$ and we get $m = \frac{10^k a}{2^4} = \frac{600 \cdot 10^{k-2}}{2^4} = 25 \cdot 10^{k-2}$. Since the last digit is not zero, we have $m = 25$, whence $n = 625$.
* If $a = 9$, then $2^k 5^k$ must be divisible by $2^3$. Therefore $k \ge 3$ and we get $m = \frac{10^k a}{2^4} = \frac{9000 \cdot 10^{k-3}}{2^4} = 375 \cdot 10^{k-3}$. Since the last digit of $m$ is not zero, we have $m = 375$, whence $n = 9375$.