This is not a Logic Problem!

Dividing both sides by $4$ gives, $5m+3n=503$ .Now,from the divisibility test of $5$ we have that $503-3n$ should either end with a $0$ or a $5$ ,as $m$ is an integer.From this we have that $n$ should either end with $1$ (to make units digit $0$ ) or $6$ (to make units digit $5$ ).Hence possibilities for $n$ are $001,006,011,016....,151,156,161,166$ ,now,each value of $n$ gives one value of $m$ ,hence we have $34$ ordered pairs.

Fairly simple:

The highest value $m$ can have is 100. Knowing this, we know that we can only reduce its value from there.

Considering this, we get a value of $n=1$ for $m=100$ .

Taking the $LCM (20,12) = 60$ , we know that the only way the equation will have a balanced pair is with negative increments of $\frac{60}{20}=3$ to $m$ .

Starting with the next lowest number, $m=99$ , we can get that: $\frac{99}{3}=33$ which means that there are 33 values for which this equation is satisfied. With this, we then know that there are 33 values between 0 and 99, and we also know that 100 is the highest.

$m=33 + 1$ hence, $m=34$

I'd like to add my solution which is a more formal approach and of course a bit longer than Adarsh Kumar 's solution:

Dividing both sides by 4 we have:

$5m+3n=503 \rightarrow 5m \equiv 503 \pmod{3} \rightarrow 2m \equiv 2 \pmod{3}$

$\rightarrow 2(m-1) \equiv 0 \pmod{3} \rightarrow gcd(3,2)=1$

$\Rightarrow m-1 \equiv 0 \pmod{3} \rightarrow m=3k+1$

Substituting in the original equation we have:

$5(3k=1)+3n=503 \rightarrow 3n=503-15k-5 \rightarrow n=166-5k > 0$

$\rightarrow 5k < 166 \rightarrow k < \lfloor \frac{166}{5} \rfloor \rightarrow 0 \le k \le 33 (\mathbb {I})$

$(\mathbb {I})$ shows that $k$ can have 34 different values and hence the answer is 34.

Dividing equation by 4 : 5m+3n = 503

We cannot reduce more : 3, 5 and 503 are coprime.

One obvious solution is (100;1) : 5x100+3x1=503

The solutions in ℤ are (100-3k;1+5k), with k €ℤ. (They are combinations between the particular solution (100;1) and the associated (-3k;+5k) to associated equation 5m+3n=0)

We want only (strictly) positives ones : 100-3k>0 and 1+5k>0. Giving -1 $\le$ k $\le$ 33. Every k in this interval gives a solution integer and strictly positive (and no duplicate) : there are 34 solutions.

20m + 12n = 2012
Simplified, 5m + 3n = 503
Letting n = 5k + 1,
503 = 3(5k + 1) + 5(100 - 3k)
0 ≤ k ≤ 100/3
0 ≤ k ≤ 33.33
n(k) = 34

Divide the equation by $4$ and find one solution $(m,\:n)=(\blue{100};\:\blue{1})$ by looking at the digits of $503$ . Then we know all integer solutions: $\begin{aligned} \red{5}m-(\red{-3})n&=503,&&&\gcd(5;\:-3)&=1&&&\Rightarrow &&&& \begin{pmatrix} m\\n \end{pmatrix}&=\begin{pmatrix} \blue{100}\\\blue{1} \end{pmatrix} + \begin{pmatrix} \red{-3}\\\red{5} \end{pmatrix}k,&k&\in\mathbb{Z} \end{aligned}$ For positive solutions, we need $m,\:n\overset{!}{>} 0$ . Check both variables separately to get $\begin{aligned} k&<\frac{100}{3}&\wedge && k&>-\frac{1}{5}&&&\Rightarrow &&&& k&\in\{0;\:\ldots;\:33\}&&&\left|\quad \boxed{34}\text{ distinct solutions!}\right. \end{aligned}$