Determine the pairs of positive integer numbers $m$ and $n$ that satisfy the equation $m^2=n^2 +m+n+2018$.
We start with the given equation:
\[ m^2 = n^2 + m + n + 2018 \]

1. Rearrange the equation to isolate the terms involving \(m\) and \(n\):
\[ m^2 - n^2 = m + n + 2018 \]

2. Recognize that \(m^2 - n^2\) can be factored using the difference of squares:
\[ (m - n)(m + n) = m + n + 2018 \]

3. Let \(a = m - n\) and \(b = m + n\). Substitute these into the equation:
\[ ab = b + 2018 \]

4. Rearrange to solve for \(a\):
\[ ab - b = 2018 \]
\[ b(a - 1) = 2018 \]

5. Since \(b(a - 1) = 2018\), we need to find pairs \((a, b)\) such that \(b\) and \(a - 1\) are factors of 2018. The prime factorization of 2018 is:
\[ 2018 = 2 \times 1009 \]

6. The factor pairs of 2018 are:
\[ (1, 2018), (2, 1009), (1009, 2), (2018, 1) \]

7. For each factor pair \((a - 1, b)\), we solve for \(a\) and \(b\):
   - For \(a - 1 = 1\) and \(b = 2018\):
     \[ a = 2 \]
     \[ b = 2018 \]
     \[ m + n = 2018 \]
     \[ m - n = 2 \]

     Solving these simultaneous equations:
     \[ m + n = 2018 \]
     \[ m - n = 2 \]

     Adding these equations:
     \[ 2m = 2020 \]
     \[ m = 1010 \]

     Subtracting the second from the first:
     \[ 2n = 2016 \]
     \[ n = 1008 \]

     Thus, one solution is \((m, n) = (1010, 1008)\).

   - For \(a - 1 = 2\) and \(b = 1009\):
     \[ a = 3 \]
     \[ b = 1009 \]
     \[ m + n = 1009 \]
     \[ m - n = 3 \]

     Solving these simultaneous equations:
     \[ m + n = 1009 \]
     \[ m - n = 3 \]

     Adding these equations:
     \[ 2m = 1012 \]
     \[ m = 506 \]

     Subtracting the second from the first:
     \[ 2n = 1006 \]
     \[ n = 503 \]

     Thus, another solution is \((m, n) = (506, 503)\).

8. The other factor pairs \((1009, 2)\) and \((2018, 1)\) do not yield new solutions as they are symmetric to the ones already found.

The final answer is \( \boxed{ (m, n) = (1010, 1008) } \) and \((m, n) = (506, 503)\).