Answer: 34

Small Hint:

Divide the equation by $4$ to get $5m + 3n = 503$

Big Hint:

Modulo $3$ the equation forces $m \equiv 1 \pmod{3};$ count the values $m = 3k + 1$ that keep $n$ positive

Solution:

Dividing by $4$ gives $5m + 3n = 503.$ Reducing modulo $3,$ we need $2m \equiv 503 \equiv 2 \pmod{3},$ so $m \equiv 1 \pmod{3}.$ Write $m = 3k + 1$ with $k \ge 0;$ then $3n = 503 - 5(3k + 1)$ $= 498 - 15k,$ so $n = 166 - 5k.$

This is positive exactly when $5k \le 165,$ that is $k \le 33.$ So $k = 0, 1, \ldots, 33$ all work, giving $34$ ordered pairs.