Problem 1884

$k = \lfloor 2024/113 \rfloor = 17$. In general, for $n$ robots and $m$ numbers, $k = \lfloor n/m \rfloor$.

Upper bound estimate: Number the robots $1$ through $n$, let the set of numbers be $C = \{0, 1, \dots, m-1\}$, the number worn by robot $i$ be $x_i \in C$, and the program be
$$f_i(x_1, x_2, \dots, x_{i-1}, x_{i+1}, \dots, x_n) : C^n \to C.$$
Suppose Dee Dee randomly writes $X_i$ on robot $i$, where $X_i$ follows a uniform distribution on $C$, independent of the other $X_j$. Then by independence, it is easy to see that
$$E[1_{\{f_i=X_i\}}] = P\{f_i(X_1, X_2, \dots, X_{i-1}, X_{i+1}, \dots, X_n) = X_i\} = \frac{1}{m}.$$
This means the expected number of robots who guess correctly $= \sum E[1_{\{f_i=X_i\}}] = n/m$, and this also implies that Dee Dee must have a way of writing the numbers such that the number of robots who guess correctly is at most $\lfloor n/m \rfloor$.

Construction: Consider
$$f_i = i - \sum_{j \neq i} x_j \quad \bmod m$$
Note that robot $i$ guesses correctly if and only if
$$x_i \equiv i - \sum_{j \neq i} x_j \quad \bmod m \Leftrightarrow \sum_j x_j \equiv i \quad \bmod m, \quad (1)$$
and there must be at least $\lfloor n/m \rfloor$ values of $i$ for which the last expression of (1) holds, so the result follows.