Let's assume the agents are labeled $[0, n)$. Let $w > 2$ be a small constant, e.g. 4. Let $F_k(x)$ be an oblivious pseudo-random function (OPRF).
Let each agent generate random value $s_i$, announce $H(s_i)$ and then choose $r = \bigoplus_i s_i$ to ensure a fair random shared seed (peers checking the hashes).
Initially agent $i$ chooses $X_i = i$. Then, some rounds take place. Use the chosen seed to derive a different $P_r$ and $w$ agents $A_{r,0}, A_{r,1}, \cdots, A_{r,w-1}$ for each round. Then, in round $r$ every agent computes
$$X'_i = (P_r - X_i) \bmod n$$
$$\hat X_i = \max(X_i, X'_i)$$
$$p = H(\hat X_i) \bmod w$$
$$b =\text{get-secret}\left(p, r, \hat{X_i}\right) \bmod 2$$
where $\text{get-secret}(p, r, \hat{X})$ securely contacts agent $A_{r,p}$ and uses the OPRF primitive to compute $F_{K_{p,r}}(\hat{X})$, where $K_{p,r}$ is some never-shared secret round key generated (and stored) by agent $A_{r,p}$. Then, at the end of the round iff $b = 1$ the agent replaces $X_i$ with $X'_i$.
After the round is over the $w$ chosen agents $A$ can (and should) communicate to ensure no one cheated and asked for more than one secret. This ensures everyone only has partial information. If you force the $\text{get-secret}$ requests to be signed by agent $i$ one can even provide evidence another agent tried to cheat.
Do this for a sufficient* number of rounds and the result is that each agent $i$ has a value $X_i$ such that there is a permutation $\sigma(i) = X_i$, yet every agent only knows their own $\sigma(i)$, and is deeply unsure about anyone else's $\sigma(j)$.
However, this just gives us a secret random permutation. But we can use this to get a pairing. Note that if we let $g(i) = i \oplus 1$ (where $\oplus$ is XOR) then $f(i) = \sigma^{-1}(g(i))$ is a pairing (assuming zero-indexed permutations).
How can we compute $\sigma^{-1}$? By noting that each round is it's own inverse, each agent sets $X_i = g(X_i)$, and we run all the rounds again, in reverse order (with the same $P,A$ and $K$ values), again ensuring that no agent requests too many secrets. The end result is that $X_i$ contains the partner for each agent.
* The technique above is inspired by sometimes recurse shuffle. Read the paper for more ideas, and to get some bounds for what is sufficient.
