Let 16 students take part in a competition where each problem is multiple choice with four choices. We are to find the maximum number of problems such that any two students have at most one answer in common.

Let $T$ denote the number of triples $(S_i, S_j, Q_k)$ such that students $S_i$ and $S_j$ answered the same for question $Q_k$.

First, consider the number of ways to choose 2 students out of 16, which is given by:
$$\binom{16}{2} = \frac{16 \times 15}{2} = 120.$$

Since any two students have at most one answer in common, we have:
$$T \leq \binom{16}{2} = 120.$$

Next, let $x, y, z, w$ be the number of students choosing the first, second, third, and fourth options respectively for a given question, and let there be $m$ questions in total.

Applying the lemma and using the Cauchy-Schwarz inequality, we get:
$$\binom{x}{2} + \binom{y}{2} + \binom{z}{2} + \binom{w}{2} \geq 24.$$

Thus, for $m$ questions, we have:
$$T \geq 24m.$$

Combining the inequalities, we get:
$$24m \leq 120 \implies m \leq 5.$$

Therefore, the maximum number of problems is:
$\text{[ 5 .}$