Problem 1275

1. Understanding the Problem:
We need to find the sum $a_1 + a_2 + \ldots + a_{p-1}$, where $a_k$ is the number of divisors of $kp + 1$ that lie strictly between $k$ and $p$.

2. Analyzing the Divisors:
For each $k$ in the range $1 \le k \le p-1$, consider the number $kp + 1$. We need to count the divisors of $kp + 1$ that are strictly between $k$ and $p$.

3. Claim:
For a fixed $m$ (where $1 < m < p$), $m$ contributes exactly once to one of $a_1, a_2, \ldots, a_{m-1}$.

4. Proof of the Claim:
Consider the numbers $p+1, 2p+1, \ldots, (m-1)p+1$. We need to show that $m$ divides exactly one of these numbers.

- Since $\gcd(m, p) = 1$ (because $p$ is a prime and $m < p$), $m$ does not divide $p$.
- Therefore, $m$ must divide $kp + 1$ for some $k$ in the range $1 \le k \le m-1$.

To see why, consider the congruences:
$$p+1 \equiv 1 \pmod{m}, \quad 2p+1 \equiv 2 \pmod{m}, \quad \ldots, \quad (m-1)p+1 \equiv m-1 \pmod{m}$$
Since $\gcd(m, p) = 1$, the numbers $p, 2p, \ldots, (m-1)p$ form a complete residue system modulo $m$. Thus, $kp + 1$ will be congruent to $1, 2, \ldots, m-1$ modulo $m$, respectively.

Therefore, $m$ will divide exactly one of these numbers $kp + 1$ for $k$ in the range $1 \le k \le m-1$.

5. Summing Up Contributions:
Each $m$ (where $1 < m < p$) contributes exactly once to one of $a_1, a_2, \ldots, a_{m-1}$. Since there are $p-2$ such $m$ (from $2$ to $p-1$), the total sum $a_1 + a_2 + \ldots + a_{p-1}$ is $p-2$.

The final answer is $\boxed{p-2}$.