(2N - 3) telephone calls, for N = 2,3 

(2N - 4) telephone calls, for N > 3 

Divide the N secret agents into two groups. If N is odd, one
group will contain one extra agent. 

Consider first group: agent 1 will call up agent 2, agent 2
will call up agent 3 and so on. Similarly in second group,
agent 1 will call up agent 2, agent 2 will call up agent 3
and so on. After (N - 2) calls, two agents in each the group
will know anything that anyone knew in his group, say they
are Y1 & Y2 from group 1 and Z1 & Z2 from group 2. 

Now, Y1 will call up Z1 and Y2 will call up Z2. Hence, in
next two calls total of 4 agents will know everything. 

Now (N - 4) telephone calls are reqiured for remaining (N -
4) secret agents. 

Total telephone calls require are 

= (N - 2) + 2 + (N - 4) 

= 2N - 4 

Let\'s take an example. Say there are 4 secret agents W, X,
Y & Z. Divide them into two groups of 2 each i.e. (W, X) and
(Y, Z). Here, 4 telephone calls are required. 

1.	W will call up X. 

2.	Y will call up Z. 

3.	W, who knows WX will call up Y, who knows YZ. 

4.	X, who knows WX will call up Z, who knows YZ. 

Take an another example. Say there are 5 secret agents J, K,
L, M & N. Divide them into two groups i.e. (J, K) and (L, M,
N). Here, 6 telephone calls are required. 

1.	J will call up K. 

2.	L will call up M. 

3.	M will call up N. Now M and N know LMN. 

4.	J, who knows JK will call up M, who knows LMN. 

5.	K, who knows JK will call up N, who knows LMN. 

6.	L will call up to anyone of four.