Ireland

We represent the given information by a graph the vertices of which correspond to the $117$ spies. Two vertices are connected iff the corresponding spies share a password which allows them to communicate directly. We need to ensure that at least one of the sub-graphs formed by spies sharing the same mission is connected.

To find the number of edges needed to ensure that a graph with $a\ge 2$ vertices is connected, no matter how the edges are placed, we first observe that the complete graph with $a-1$ vertices has $\frac{(a-1)(a-2)}{2}$ edges. We next prove that it is sufficient to have $$\frac{(a-1)(a-2)}{2}+1=\frac{a^2-3a+4}{2}$$ edges. To see this, assume that the graph is not connected. Then there exist two sub-graphs, not connected to each other, one with $x\ge 1$ and the other with $a-x\ge 1$ vertices. The number of edges in such a graph is at most $$\begin{array}{rl}\frac{(a-x)(a-x-1)}{2}+\frac{x(x-1)}{2}& =\frac{2x^2-2ax+a^2-a}{2}=\\ & =\frac{a^2-3a+2}{2}-(a-x-1)(x-1)\le \frac{a^2-3a+2}{2}=\frac{(a-1)(a-2)}{2}\end{array}$$

If the sizes of the three missions are $a$, $b$ and $c$, the above shows that issuing $$n=\frac{a^2-3a+2}{2}+\frac{b^2-3b+2}{2}+\frac{c^2-3c+2}{2}+1$$ passwords, ensures that at least one mission is networked. Finally, we need to minimise $2n=a^2+b^2+c^2-3(a+b+c)+8$ subject to the condition $a+b+c=117$. Because $$\begin{array}{rl}3(a^2+b^2+c^2)-(a+b+c{)}^2& =2(a^2+b^2+c^2-ab-bc-ca)=\\ & =(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ge 0,\end{array}$$ we obtain $$2n\ge \frac{(a+b+c{)}^2}{3}-3(a+b+c)+8=\frac{117^2}{3}-3\cdot 117+8=4220,$$ with equality iff $a=b=c=39$. This shows that the minimum number of passwords needed to ensure that at least one mission is networked is $n=2110$. This number is sufficient iff the three missions are of equal size.