Suppose that your team contains $n$ people. Each round of discussions ("session") corresponds to a matching in the complete graph $K_n$. Therefore you want to partition into as few matchings as possible. The answer depends on the parity of $n$.
Case 1: $n$ is even. In this case, $K_n$ contains $\binom{n}{2}$ edges, and a matching contains at most $n/2$ edges. Hence we can hope for a partition of $K_n$ into $n-1$ perfect matchings. This is a standard problem, known as 1-factorization of the complete graph. Here is one solution:
$$
(0,n-1),(1,n-2),(2,n-3),(3,n-4),\ldots,(n/2-1,n/2) \\
(1,n-1),(2,0),(3,n-2),(4,n-3),\ldots,(n/2,n/2+1) \\
(2,n-1),(3,1),(4,0),(5,n-2),\ldots,(n/2+1,n/2+2) \\
\ldots
$$
Explanation: the $i$'th row consists of $(i,n-1)$ together with $(j+i,n-1-j+i \bmod n-1)$ for $1 \leq j \leq n/2-1$. For example, when $n = 6$ you get:
$$
(0,5),(1,4),(2,3) \\
(1,5),(2,0),(3,4) \\
(2,5),(3,1),(4,0) \\
(3,5),(4,2),(0,1) \\
(4,5),(0,3),(1,2)
$$
Case 2: $n$ is odd. In this case, $K_n$ still contains $\binom{n}{2}$ edges, but now each matching contains at most $(n-1)/2$ edges. Hence we can hope for a partition of $K_n$ into $n$ almost-perfect matchings. We can take the solution above for $n+1$, and just ignore the match of vertex $n$. This gives the following solution:
$$
(1,n-1),(2,n-2),(3,n-3),\ldots,((n-1)/2,(n+1)/2) \\
(2,0),(3,n-1),(4,n-2),\ldots,((n+1)/2,(n+3)/2) \\
(3,1),(4,0),(5,n-1),\ldots,((n+3)/2,(n+5)/2) \\
\ldots
$$
Explanation: the $i$th row consists of $(j+i,n-j+i \bmod n)$ for $1 \leq j \leq (n-1)/2$. For example, when $n = 7$ you get:
$$
(1,6),(2,5),(3,4) \\
(2,0),(3,6),(4,5) \\
(3,1),(4,0),(5,6) \\
(4,2),(5,1),(6,0) \\
(5,3),(6,2),(0,1) \\
(6,4),(0,3),(1,2) \\
(0,5),(1,4),(2,3)
$$
