Suppose that $a$ is real number. Sequence $a_1,a_2,a_3,....$ satisfies
$$a_1=a, a_{n+1} = 
\begin{cases}
a_n - \frac{1}{a_n}, & a_n\ne 0 \\
0, & a_n=0
\end{cases}
(n=1,2,3,..)$$
Find all possible values of $a$ such that $|a_n|<1$ for all positive integer $n$.

1. **Initial Condition and Function Definition**:
   Given the sequence \(a_1, a_2, a_3, \ldots\) defined by:
   \[
   a_1 = a, \quad a_{n+1} = 
   \begin{cases}
   a_n - \frac{1}{a_n}, & a_n \ne 0 \\
   0, & a_n = 0
   \end{cases}
   \]
   we need to find all possible values of \(a\) such that \(|a_n| < 1\) for all positive integers \(n\).

2. **Case \(a = 0\)**:
   If \(a = 0\), then \(a_1 = 0\) and \(a_n = 0\) for all \(n\). Clearly, \(|a_n| = 0 < 1\) for all \(n\). Thus, \(a = 0\) is a valid solution.

3. **General Case \(a \ne 0\)**:
   Define the function \(f(x) = x - \frac{1}{x}\). We need to analyze the behavior of the sequence generated by this function.

4. **Behavior of \(f(x)\)**:
   Notice that:
   \[
   f(-x) = -x - \frac{1}{-x} = -\left(x - \frac{1}{x}\right) = -f(x)
   \]
   and
   \[
   f\left(\frac{1}{x}\right) = \frac{1}{x} - x
   \]
   This symmetry suggests that we only need to consider \(a \in (-1, 1)\).

5. **Fixed Points of \(f(x)\)**:
   To find the fixed points, solve \(f(x) = x\):
   \[
   x - \frac{1}{x} = x \implies -\frac{1}{x} = 0
   \]
   This equation has no real solutions. However, consider the modified function \(g(x) = \frac{1}{x} - x\):
   \[
   g(x) = x \implies \frac{1}{x} - x = x \implies \frac{1}{x} = 2x \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}
   \]
   Thus, the fixed points are \(x = \pm \frac{\sqrt{2}}{2}\).

6. **Stability of Fixed Points**:
   Calculate the derivative of \(f(x)\):
   \[
   f'(x) = 1 + \frac{1}{x^2}
   \]
   At the fixed points \(x = \pm \frac{\sqrt{2}}{2}\):
   \[
   f'\left(\pm \frac{\sqrt{2}}{2}\right) = 1 + \left(\frac{\sqrt{2}}{2}\right)^{-2} = 1 + 2 = 3
   \]
   Since \(|f'(x)| > 1\), the fixed points are repelling.

7. **Conclusion**:
   Since the fixed points are repelling, any initial value \(a\) close to \(\pm \frac{\sqrt{2}}{2}\) will eventually lead to \(|a_n| \ge 1\). Therefore, the only value of \(a\) that ensures \(|a_n| < 1\) for all \(n\) is \(a = 0\).

The final answer is \(\boxed{0}\)