How many integers $n$ are there such that $(n+1!)(n+2!)(n+3!)\cdots(n+2013!)$ is divisible by $210$ and $1 \le n \le 210$?

[i]Proposed by Lewis Chen[/i]
To determine how many integers \( n \) satisfy the condition that \((n+1!)(n+2!)(n+3!)\cdots(n+2013!)\) is divisible by \( 210 \) and \( 1 \le n \le 210 \), we need to analyze the prime factorization of \( 210 \). 

The prime factorization of \( 210 \) is:
\[ 210 = 2 \times 3 \times 5 \times 7 \]

For the product \((n+1!)(n+2!)(n+3!)\cdots(n+2013!)\) to be divisible by \( 210 \), it must be divisible by each of the primes \( 2, 3, 5, \) and \( 7 \). 

1. **Divisibility by 2:**
   - Since \( n+1! \) is part of the product, and \( 1! = 1 \), \( n+1 \) must be even. Therefore, \( n \) must be odd.

2. **Divisibility by 3:**
   - Since \( n+2! \) is part of the product, and \( 2! = 2 \), \( n+2 \) must be divisible by 3. Therefore, \( n \equiv 1 \pmod{3} \).

3. **Divisibility by 5:**
   - Since \( n+4! \) is part of the product, and \( 4! = 24 \), \( n+24 \) must be divisible by 5. Therefore, \( n \equiv 1 \pmod{5} \).

4. **Divisibility by 7:**
   - Since \( n+6! \) is part of the product, and \( 6! = 720 \), \( n+720 \) must be divisible by 7. Therefore, \( n \equiv 1 \pmod{7} \).

Next, we use the Chinese Remainder Theorem (CRT) to find the number of solutions \( n \) that satisfy these congruences simultaneously.

5. **Applying the Chinese Remainder Theorem:**
   - We have the system of congruences:
     \[
     \begin{cases}
     n \equiv 1 \pmod{2} \\
     n \equiv 1 \pmod{3} \\
     n \equiv 1 \pmod{5} \\
     n \equiv 1 \pmod{7}
     \end{cases}
     \]
   - Since \( 2, 3, 5, \) and \( 7 \) are pairwise coprime, by the Chinese Remainder Theorem, there is a unique solution modulo \( 2 \times 3 \times 5 \times 7 = 210 \).

6. **Counting the solutions:**
   - The unique solution modulo \( 210 \) is \( n \equiv 1 \pmod{210} \).
   - Since \( 1 \le n \le 210 \), the only solution is \( n = 1 \).

Therefore, there is exactly one integer \( n \) in the range \( 1 \le n \le 210 \) that satisfies the given condition.

The final answer is \(\boxed{1}\)