If $1<k_1<k_2<...<k_n$ and $a_1,a_2,...,a_n$ are integers such that for every integer $N,$ $k_i \mid N-a_i$ for some $1 \leq i \leq n,$ find the smallest possible value of $n.$
To solve the problem, we need to find the smallest possible value of \( n \) such that for every integer \( N \), there exists an integer \( k_i \) from the set \( \{k_1, k_2, \ldots, k_n\} \) that divides \( N - a_i \) for some \( 1 \leq i \leq n \). 

1. **Initial Assumptions and Setup:**
   - Given \( 1 < k_1 < k_2 < \ldots < k_n \) and \( a_1, a_2, \ldots, a_n \) are integers.
   - For every integer \( N \), there exists \( k_i \) such that \( k_i \mid (N - a_i) \).

2. **Checking for \( n = 5 \):**
   - Consider the integers \( k_1 = 2, k_2 = 3, k_3 = 4, k_4 = 6, k_5 = 12 \).
   - Every integer \( N \) is congruent to one of \( 0 \pmod{2}, 1 \pmod{3}, 3 \pmod{4}, 5 \pmod{6}, 9 \pmod{12} \).
   - We need to check if these congruences cover all residue classes modulo 12.

3. **Verification for \( n = 5 \):**
   - We need to check if every integer \( N \) modulo 12 can be represented by one of the given congruences.
   - The residue classes modulo 12 are \( \{0, 1, 2, \ldots, 11\} \).
   - By checking each residue class, we find that they are indeed covered by the given congruences.

4. **Proving \( n = 4 \) is Impossible:**
   - Suppose \( n = 4 \). We need to show that it is impossible to cover all integers \( N \) with only 4 congruences.
   - We can choose \( 0 \leq a_i < k_i \).
   - For any \( l > k_n \), the number of integers \( \equiv a_i \pmod{k_i} \) is \( \left\lfloor \frac{l-1-a_i}{k_i} \right\rfloor \), which is less than \( \frac{l}{k_i} \).
   - Therefore, \( l \leq \sum_{i=1}^{n} \left\lfloor \frac{l-1-a_i}{k_i} \right\rfloor < \sum_{i=1}^{n} \frac{l}{k_i} \).
   - This implies \( 1 < \sum_{i=1}^{n} \frac{1}{k_i} \).

5. **Summing the Reciprocals:**
   - For \( n = 4 \), we need \( \sum_{i=1}^{4} \frac{1}{k_i} > 1 \).
   - Consider \( k_1 = 2 \). Then \( \frac{1}{2} + \frac{1}{k_2} + \frac{1}{k_3} + \frac{1}{k_4} > 1 \).
   - If \( k_2, k_3, k_4 \) are powers of 2, \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} < 1 \).
   - Therefore, at least one \( k_i \) must be an odd prime.

6. **Considering Odd Primes:**
   - Suppose \( p \) divides \( k_i \) but not the other \( k_j, k_s \) with \( j, s > 1 \).
   - By the Chinese Remainder Theorem, \( N \equiv a \pmod{2} \) and \( N \equiv b \pmod{p} \) implies \( N \equiv c \pmod{2p} \).
   - For \( N \equiv c \pmod{2p} \), \( N \not\equiv a_i \pmod{k_i} \) and \( N \not\equiv a_1 \pmod{2} \).
   - This implies \( k_j, k_s \) must divide \( 4p \), leading to contradictions.

7. **Final Contradiction:**
   - If \( p \) divides all \( k_2, k_3, k_4 \), and \( p \geq 5 \), then \( \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{15} < 1 \).
   - If \( p = 3 \), then \( k_2 = 3 \), and similar contradictions arise.

8. **Conclusion:**
   - Therefore, \( n = 4 \) is impossible, and the smallest possible value of \( n \) is \( 5 \).

The final answer is \( \boxed{5} \).