Problem 1793

1. Define Good Integers:
Let $n$ be a positive integer such that $p \mid n! + 1$. We call such an integer $n$ a good integer. Note that $p-1$ is the largest good integer because for $n \geq p$, $n!$ is divisible by $p$, making $n! + 1$ not divisible by $p$.

2. Lemma 1:
For each positive integer $t$, there are at most $t$ good integers $k$ such that $k + t$ is also good.

Proof of Lemma 1:
Suppose $p$ divides both $(k+t)! + 1$ and $k! + 1$. Then:
$$p \mid \big((k+t)! + 1\big) - \big(k! + 1\big) = k! \big((k+1)(k+2) \cdots (k+t) - 1\big).$$
Since $k$ is a good integer, $\gcd(p, k!) = 1$. Therefore:
$$p \mid (k+1)(k+2) \cdots (k+t) - 1.$$
The right-hand side is a polynomial of degree $t$ in $k$, which has at most $t$ solutions in $\mathbb{F}_p$ by Lagrange's theorem. $\blacksquare$

3. Sequence of Good Integers:
Let $n_1 < n_2 < \cdots < n_m = p-1$ denote the set of good integers. For each $1 \leq j \leq p$, let $a_j$ denote the number of integers $1 \leq k \leq m$ such that $n_k - n_{k-1} = j$. By Lemma 1, $a_j \leq j$ for each $j$.

4. Sum of Differences:
We have:
$$a_1 + 2a_2 + \cdots + pa_p = \sum_{k=1}^m (n_k - n_{k-1}) = n_m - n_1 \leq p.$$
We wish to maximize $a_1 + a_2 + \cdots + a_p$.

5. Lemma 2:
Suppose $b_1, \ldots, b_n$ are nonnegative real numbers such that:
$$b_1 + 2b_2 + \cdots + nb_n \leq n \quad \text{and} \quad b_i \leq i \text{ for all } i.$$
Then there exists a constant $c > 0$ such that:
$$b_1 + b_2 + \cdots + b_n \leq c n^{2/3}.$$

Proof of Lemma 2:
Without loss of generality, we may replace the condition with $b_1 + 2b_2 + \cdots + nb_n = n$. Consider an arbitrary $n$-tuple $(b_1, \ldots, b_n)$ of nonnegative real numbers satisfying the given conditions. If there exist integers $r < s$ with $b_r < r$ but $b_s > 0$, then we may replace $(b_1, \ldots, b_n)$ by the tuple:
$$(b_1', \ldots, b_n') = (b_1, \ldots, b_{r-1}, b_r + \frac{s\varepsilon}{r}, b_{r+1}, \ldots, b_{s-1}, b_s - \varepsilon, b_{s+1}, \ldots),$$
where $\varepsilon$ is chosen small enough that $b_r + \frac{s\varepsilon}{r} < r$ and $b_s > \varepsilon$. Then:
$$b_1' + 2b_2' + \cdots + nb_n' = b_1 + 2b_2 + \cdots + nb_n = n,$$
but the sum $b_1 + b_2 + \cdots + b_n$ increases. This means that $(b_1, \ldots, b_n)$ does not achieve our maximum. In particular, our maximum is achieved when:
$$b_j = \begin{cases} j & \text{if } 1 \leq j \leq t, \\ 0 & \text{if } j \geq t+2, \end{cases}$$
where $t$ is some positive integer at most $n$.

Finally, observe that:
$$n = b_1 + 2b_2 + \cdots + nb_n \geq 1^2 + 2^2 + \cdots + t^2 \gtrsim t^3,$$
and hence $t \lesssim n^{1/3}$. Then:
$$b_1 + b_2 + \cdots + b_n = b_1 + \cdots + b_{t+1} \leq 1 + \cdots + (t+1) \lesssim t^2 \lesssim n^{2/3},$$
and so $b_1 + \cdots + b_n \leq c n^{2/3}$ for some constant $c$. $\blacksquare$

6. Application of Lemma 2:
Apply Lemma 2 to the sequence $(a_1, \ldots, a_p)$. Since $a_1 + 2a_2 + \cdots + pa_p \leq p$ and $a_i \leq i$, we conclude that:
$$a_1 + a_2 + \cdots + a_p \leq c p^{2/3}.$$

The final answer is $\boxed{c p^{2/3}}$.