To find all pairs $(n, p)$ of positive integers such that $p$ is prime and:

$$1 + 2 + \cdots + n = 3 \cdot (1^2 + 2^2 + \cdots + p^2)$$

we first express these sums using known formulas:

1. The sum of the first $n$ positive integers is given by:
$$\frac{n(n + 1)}{2}$$

2. The sum of the squares of the first $p$ positive integers is given by:
$$\frac{p(p + 1)(2p + 1)}{6}$$

Substituting these into the equation, we have:

$$\frac{n(n + 1)}{2} = 3 \cdot \frac{p(p + 1)(2p + 1)}{6}$$

Simplifying the right-hand side:

$$\frac{n(n + 1)}{2} = \frac{p(p + 1)(2p + 1)}{2}$$

Thus, we have:

$$n(n + 1) = p(p + 1)(2p + 1)$$

Now, let's solve for $(n, p)$. We test small values for $p$ since $p$ is a prime number.

- **Case $p = 2$:**

$$p(p + 1)(2p + 1) = 2 \cdot 3 \cdot 5 = 30$$

We need:

$$n(n + 1) = 30$$

Testing integers for $n$:

- $n = 5$ gives $n(n + 1) = 5 \cdot 6 = 30$

This is a valid solution: $(n, p) = (5, 2)$.

For larger prime numbers, the product $p(p + 1)(2p + 1)$ grows quickly, making $n(n + 1) = p(p + 1)(2p + 1)$ impractical for small $n$. Therefore, no other small primes will provide solutions where $n$ is a reasonable positive integer.

Thus, the only solution to the problem is:

$$\boxed{(5, 2)}$$