Difficulty of a congruence problem
Combining fgrieu's and my observation, it appears that a large $r$ can be recovered given enough $p_iq_i + r$ values.

In particular, we will assume that both $p_i$ and $q_i$ are selected uniformly.

Then, $(p_iq_i + r) = r \pmod s$ will hold will probability at least $(2s-1)/s^2 \approx 2/s$ (as it will hold if either $p_i$ or $q_i$ is 0 modulo $s$, $(p_iq_i + r) \bmod s$ will be any other specific values with probability circa $1/s$.  Actually, the bias is even stronger if $s$ is a power of a smaller prime, we'll ignore that for now.

Hence, what we can do is, for various small prime-power $s$, take our $p_iq_i + r$ values, and compute $p_iq_i + r \bmod s$, and see what value happens the most often.  That'll give us the likely value of $r \bmod s$.  Then, we take all our likely values, and use the Chinese Remainder Theorem to reconstruct $r$.

If we have 10,000 such values, we can use the prime-power's up to 710; for each value of $s$, the correct value of $r \bmod s$ will be at least 3 standard deviations over the rest (and hence the most common value is likely to be the correct one), and that'll give us enough relations to recover a 1024 bit $r$.

We can then use fgrieu's observation to validate the $r$ we recovered.