Find all the pairs of positive integers $(m,n)$ such that the numbers $A=n^2+2mn+3m^2+3n$, $B=2n^2+3mn+m^2$, $C=3n^2+mn+2m^2$ are consecutive in some order.
We need to find all pairs of positive integers \((m, n)\) such that the numbers \(A = n^2 + 2mn + 3m^2 + 3n\), \(B = 2n^2 + 3mn + m^2\), and \(C = 3n^2 + mn + 2m^2\) are consecutive in some order.

First, let's consider the differences between these expressions:

1. Calculate \(C - B\):
\[
C - B = (3n^2 + mn + 2m^2) - (2n^2 + 3mn + m^2) = n^2 - 2mn + m^2 = (n - m)^2
\]
Since \((n - m)^2 \geq 0\), we have \(C \geq B\). For \(C\) and \(B\) to be consecutive, \((n - m)^2\) must be 1:
\[
(n - m)^2 = 1 \implies n - m = \pm 1 \implies n = m \pm 1
\]

2. Consider the case \(n = m - 1\):
\[
A = (m-1)^2 + 2m(m-1) + 3m^2 + 3(m-1) = m^2 - 2m + 1 + 2m^2 - 2m + 3m^2 + 3m - 3 = 6m^2 - m - 2
\]
\[
B = 2(m-1)^2 + 3m(m-1) + m^2 = 2(m^2 - 2m + 1) + 3m^2 - 3m + m^2 = 2m^2 - 4m + 2 + 3m^2 - 3m + m^2 = 6m^2 - 7m + 2
\]
\[
C = 3(m-1)^2 + m(m-1) + 2m^2 = 3(m^2 - 2m + 1) + m^2 - m + 2m^2 = 3m^2 - 6m + 3 + m^2 - m + 2m^2 = 6m^2 - 7m + 3
\]

Here, \(A > C > B\) and \(A = C + 1\):
\[
6m^2 - m - 2 = 6m^2 - 7m + 3 + 1 \implies -m - 2 = -7m + 4 \implies 6m = 6 \implies m = 1
\]
\[
n = m - 1 = 1 - 1 = 0
\]
However, \(n\) must be a positive integer, so this case is not valid.

3. Consider the case \(n = m + 1\):
\[
A = (m+1)^2 + 2m(m+1) + 3m^2 + 3(m+1) = m^2 + 2m + 1 + 2m^2 + 2m + 3m^2 + 3m + 3 = 6m^2 + 7m + 4
\]
\[
B = 2(m+1)^2 + 3m(m+1) + m^2 = 2(m^2 + 2m + 1) + 3m^2 + 3m + m^2 = 2m^2 + 4m + 2 + 3m^2 + 3m + m^2 = 6m^2 + 7m + 2
\]
\[
C = 3(m+1)^2 + m(m+1) + 2m^2 = 3(m^2 + 2m + 1) + m^2 + m + 2m^2 = 3m^2 + 6m + 3 + m^2 + m + 2m^2 = 6m^2 + 7m + 3
\]

Here, \(A > C > B\) and \(A = C + 1\), \(C = B + 1\), and \(A = B + 2\). This condition holds for every \(m\) such that \(n = m + 1\).

Thus, the solutions are \((m, n) = (k, k+1)\) where \(k\) is a positive integer.

The final answer is \( \boxed{ (m, n) = (k, k+1) } \) where \(k\) is a positive integer.