Let $\{a_i\}_{i=0}^\infty$ be a sequence of real numbers such that \[\sum_{n=1}^\infty\dfrac {x^n}{1-x^n}=a_0+a_1x+a_2x^2+a_3x^3+\cdots\] for all $|x|<1$.  Find $a_{1000}$.

[i]Proposed by David Altizio[/i]
1. We start with the given series:
   \[
   \sum_{n=1}^\infty \frac{x^n}{1-x^n} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots
   \]
   We need to find the coefficient \(a_{1000}\).

2. Recall the generating function for a geometric series:
   \[
   \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots
   \]
   Substituting \(x^n\) for \(x\), we get:
   \[
   \frac{1}{1-x^n} = 1 + x^n + x^{2n} + x^{3n} + \cdots
   \]
   Multiplying both sides by \(x^n\), we obtain:
   \[
   \frac{x^n}{1-x^n} = x^n + x^{2n} + x^{3n} + x^{4n} + \cdots
   \]

3. The series \(\sum_{n=1}^\infty \frac{x^n}{1-x^n}\) can be expanded as:
   \[
   \sum_{n=1}^\infty (x^n + x^{2n} + x^{3n} + \cdots)
   \]
   This means that each term \(x^k\) in the series will appear if and only if there exists an \(n\) such that \(n \mid k\).

4. To find \(a_{1000}\), we need to count the number of divisors of 1000. The prime factorization of 1000 is:
   \[
   1000 = 2^3 \times 5^3
   \]
   The number of divisors of a number \(n = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}\) is given by:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   For 1000, this is:
   \[
   (3+1)(3+1) = 4 \times 4 = 16
   \]

5. Therefore, the coefficient \(a_{1000}\) is the number of divisors of 1000, which is 16.

The final answer is \(\boxed{16}\)