$PA_1A_2...A_n$ is a pyramid. The base $A_1A_2...A_n$ is a regular n-gon. The apex $P$ is placed so that the lines $PA_i$ all make an angle $60^{\cdot}$ with the plane of the base. For which $n$ is it possible to find $B_i$ on $PA_i$ for $i = 2, 3, ... , n$ such that $A_1B_2 + B_2B_3 + B_3B_4 + ... + B_{n-1}B_n + B_nA_1 < 2A_1P$?
1. **Understanding the Geometry**:
   - We have a pyramid \( PA_1A_2 \ldots A_n \) with a regular \( n \)-gon base \( A_1A_2 \ldots A_n \).
   - The apex \( P \) is positioned such that each line segment \( PA_i \) makes an angle of \( 60^\circ \) with the plane of the base.

2. **Positioning of Points \( B_i \)**:
   - We need to find points \( B_i \) on \( PA_i \) for \( i = 2, 3, \ldots, n \) such that the sum of the distances \( A_1B_2 + B_2B_3 + B_3B_4 + \ldots + B_{n-1}B_n + B_nA_1 \) is less than \( 2A_1P \).

3. **Analyzing the Distance Condition**:
   - Consider the projection of the pyramid onto the plane of the base. Since \( PA_i \) makes an angle of \( 60^\circ \) with the base, the height \( h \) of the pyramid from \( P \) to the base is given by:
     \[
     h = PA_i \sin(60^\circ) = \frac{\sqrt{3}}{2} PA_i
     \]
   - The length of \( PA_i \) can be denoted as \( l \). Thus, \( h = \frac{\sqrt{3}}{2} l \).

4. **Finding \( B_i \) on \( PA_i \)**:
   - Let \( B_i \) be a point on \( PA_i \) such that \( PB_i = k \cdot PA_i \) where \( 0 < k < 1 \).
   - The projection of \( B_i \) onto the base plane will be at a distance \( k \cdot PA_i \cos(60^\circ) = k \cdot \frac{1}{2} PA_i = \frac{k}{2} l \) from \( A_i \).

5. **Summing the Distances**:
   - The total distance \( A_1B_2 + B_2B_3 + \ldots + B_nA_1 \) can be approximated by the perimeter of the \( n \)-gon formed by the projections of \( B_i \) onto the base plane.
   - Each side of this \( n \)-gon is approximately \( \frac{k}{2} l \).

6. **Condition for the Sum of Distances**:
   - The perimeter of the \( n \)-gon formed by the projections of \( B_i \) is \( n \cdot \frac{k}{2} l \).
   - We need this to be less than \( 2A_1P = 2l \):
     \[
     n \cdot \frac{k}{2} l < 2l
     \]
     Simplifying, we get:
     \[
     nk < 4
     \]
     Since \( 0 < k < 1 \), the maximum value of \( k \) is 1, thus:
     \[
     n < 4
     \]

7. **Conclusion**:
   - The only integer value for \( n \) that satisfies \( n < 4 \) is \( n = 3 \).

The final answer is \( \boxed{ n = 3 } \).