We need to determine all positive integers $n$ such that
$$p(n) + p(n+4) = p(n+2) + p(n+3),$$
where $p(n)$ denotes the partition function, which counts the number of ways $n$ can be partitioned into positive integers.

To solve this, we consider the equivalent equation by setting $N = n + 4$:
$$p(N) + p(N-4) = p(N-1) + p(N-2).$$

We analyze the behavior of the partition function using coarse partitions. A partition of $n$ is called coarse if all parts are at least three. Let $q(n)$ denote the number of coarse partitions of $n$.

By generating functions, we derive the following identity for $N \geq 5$:
$$p(N) + p(N-4) - p(N-1) - p(N-2) = q(N) - q(N-3) - q(N-5) - q(N-7) - q(N-9) - \dots.$$

We then examine the cases for odd and even $N$:

1. **Odd $N \geq 5$**:
$$q(N) \leq q(N-3) + q(N-5) + q(N-7) + \dots,$$
with equality if and only if $N = 5, 7, 9$.

2. **Even $N$**:
- For $N \in \{6, 8, 10, 12\}$:
$$q(N) - 1 = q(N-3) + q(N-5) + q(N-7) + \dots.$$
- For $N \geq 14$:
$$q(N) < q(N-3) + q(N-5) + q(N-7) + \dots.$$

From the above analysis, we find that the only solutions to the equation are $N = 5, 7, 9$. Converting back to the original variable $n$ using $N = n + 4$, we obtain the solutions:
$$n = 1, 3, 5.$$

The answer is: $\boxed{1,}$ 3, 5}.