Find all numbers $N=\overline{a_1a_2\ldots a_n}$ for which $9\times \overline{a_1a_2\ldots a_n}=\overline{a_n\ldots a_2a_1}$ such that at most one of the digits $a_1,a_2,\ldots ,a_n$ is zero.
1. Let \( N = \overline{a_1a_2\ldots a_n} \) be the number we are looking for, where \( \overline{a_1a_2\ldots a_n} \) represents the digits of \( N \). We need to find \( N \) such that \( 9N = \overline{a_n\ldots a_2a_1} \), the reverse of \( N \).

2. Define \( r(N) \) as the reverse of \( N \). We are given that \( r(N) = 9N \).

3. First, consider the case where \( a_1 = 0 \). This would imply \( a_n = 0 \) because the reverse of \( N \) would also start with 0. Since at most one of the digits can be zero, \( n \) must be 1, and the only solution in this case is \( N = 0 \).

4. Now, consider the case where \( a_1 > 0 \). Since \( r(N) < 10^n \), we have:
   \[
   9N < 10^n \implies N < \frac{10^n}{9}
   \]
   This implies \( N \leq \overline{111\ldots 1} \) (with \( n \) ones).

5. Since \( N \) cannot be made of all ones (as it would not satisfy the condition \( 9N = r(N) \)), there must be a zero after some leading ones. Let \( N = \overline{1110\ldots a_n} \) with exactly \( p \) leading ones.

6. The only value \( a_n \) such that \( 9a_n \) ends with \( a_1 = 1 \) is 9. Therefore, \( a_n = 9 \).

7. If \( p > 1 \), then \( a_2 = 1 \) and \( 9N \) must begin with 99. This implies \( N \) must end with 99, but then \( 9N \) ends with 91, which contradicts the fact that \( N \) begins with 11. Therefore, \( p = 1 \).

8. Thus, \( N \) begins with 10 and \( 9N \) ends with 01, and \( N \) ends with 89. We have:
   \[
   9 \times \overline{10 b_1 \ldots b_k 89} = \overline{98 b_k \ldots b_1 01}
   \]

9. First, consider \( k = 0 \):
   \[
   9 \times 1089 = 9801
   \]
   This is a valid solution.

10. If \( k > 0 \), let \( x = \overline{b_1 \ldots b_k} \). We have:
    \[
    9x + 8 = r(x) + 8 \times 10^k
    \]
    This implies:
    \[
    x \geq \frac{8 \times 10^k - 8}{9} = \overline{888\ldots 8} \text{ (with \( k \) eights)}
    \]
    Therefore, \( b_1 \in \{8, 9\} \).

11. If \( b_1 = 8 \), then \( r(x) \) ends with 8, and \( r(x) + 8 \times 10^k \) ends with 8. This implies \( 9x + 8 \) ends with 8, so \( x \) ends with 0, which is impossible since we already used the zero in \( N \).

12. Therefore, \( b_1 = 9 \). Then \( r(x) \) ends with 9, and \( 9x + 8 \) ends with 9, so \( x \) ends with 9. 

13. By induction, \( x \) must be composed entirely of 9s. Thus, \( x = 9, 99, 999, \ldots \).

14. Hence, the solutions are:
    \[
    N = \overline{0}, \overline{1089}, \overline{10999\ldots 99989}
    \]

The final answer is \( \boxed{\overline{0}, \overline{1089}, \overline{10999\ldots 99989}} \).