Find the greatest natural number $n$ such there exist natural numbers $x_{1}, x_{2}, \ldots, x_{n}$ and natural $a_{1}< a_{2}< \ldots < a_{n-1}$ satisfying the following equations for $i =1,2,\ldots,n-1$: \[x_{1}x_{2}\ldots x_{n}= 1980 \quad \text{and}\quad x_{i}+\frac{1980}{x_{i}}= a_{i}.\]
1. **Prime Factorization of 1980**:
   First, we need to find the prime factorization of 1980.
   \[
   1980 = 2^2 \times 3^2 \times 5 \times 11
   \]
   This shows that 1980 can be expressed as a product of six prime factors (counting multiplicities).

2. **Understanding the Equations**:
   We are given the equations:
   \[
   x_1 x_2 \ldots x_n = 1980
   \]
   and
   \[
   x_i + \frac{1980}{x_i} = a_i \quad \text{for} \quad i = 1, 2, \ldots, n-1
   \]
   Here, \(a_i\) are natural numbers and \(a_1 < a_2 < \ldots < a_{n-1}\).

3. **Constraints on \(x_i\)**:
   Each \(x_i\) must be a divisor of 1980. The divisors of 1980 are:
   \[
   1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20, 22, 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 180, 198, 220, 330, 396, 495, 660, 990, 1980
   \]

4. **Maximizing \(n\)**:
   To maximize \(n\), we need to distribute the prime factors among the \(x_i\) such that the product of all \(x_i\) equals 1980. Since 1980 has six prime factors, the maximum number of \(x_i\) we can have is six, provided we can find a set of six divisors whose product is 1980.

5. **Possible Set of \(x_i\)**:
   One possible set of \(x_i\) that satisfies the product condition is:
   \[
   \{11, 6, 5, 3, 2, 1\}
   \]
   Let's verify:
   \[
   11 \times 6 \times 5 \times 3 \times 2 \times 1 = 1980
   \]
   This confirms that the product is indeed 1980.

6. **Checking the Equations**:
   We need to check if the equations \(x_i + \frac{1980}{x_i} = a_i\) hold for \(i = 1, 2, \ldots, n-1\):
   \[
   \begin{align*}
   x_1 = 11 & \implies 11 + \frac{1980}{11} = 11 + 180 = 191 \\
   x_2 = 6 & \implies 6 + \frac{1980}{6} = 6 + 330 = 336 \\
   x_3 = 5 & \implies 5 + \frac{1980}{5} = 5 + 396 = 401 \\
   x_4 = 3 & \implies 3 + \frac{1980}{3} = 3 + 660 = 663 \\
   x_5 = 2 & \implies 2 + \frac{1980}{2} = 2 + 990 = 992 \\
   \end{align*}
   \]
   The values \(a_i\) are \(191, 336, 401, 663, 992\), which are in increasing order.

Conclusion:
The maximum number \(n\) such that there exist natural numbers \(x_1, x_2, \ldots, x_n\) and natural \(a_1 < a_2 < \ldots < a_{n-1}\) satisfying the given conditions is \(6\).

The final answer is \(\boxed{6}\).