If you define $S(n)$ as your sum:
$S(n)=\sum_{m=1}^n \frac{m}{(m,n)}$
then we will show that $2S(n)-1$ is multiplicative.
You can rewrite the sum as:
$S(n)=\sum_{d\mid n} \sum_{m=1}_{(m,n)=d}^n \frac{m}{d}$
Setting $k=m/d$, we get:
$S(n)=\sum_{d\mid n} \sum_{k=1}_{(k,n/d)=1}^{n/d} k = \sum_{d\mid n}\sum_{k=1}_{(k,d)=1}^d k$
So let $f(d) = \sum_{k=1}_{(k,d)=1}^d k$
It is not hard to show that this $f(1)=1$ and $f(d)=\frac{1}{2}\phi(d)d$ when $d>1$.
So we get:
$S(n)=\frac{1}{2}+\frac{1}{2}\sum_{d\mid n} \phi(d)d$
Now $g(n)=\sum_{d|n} d\phi(d)$ is multiplicative, since $n\phi(n)$ is, so $g(n)$ is determined by $g(p^a) = 1+\sum_{i=1}^a p^i\phi(p^i) = \\1+\sum_{i=1}^a p^{2i-1}(p-1) = \\1 + p(p-1)\sum_{i=1}^a p^{2i-2} = \\1+p(p-1)\frac{p^{2a}-1}{p^2-1}=\\ \frac{p^{2a+1}+1}{p+1}$
And in general, if $n=\prod p_i^{a_i}$ then $g(n)=\prod \frac{p_i^{2a_i+1}+1}{p_i+1}$
And your original sum is:
$S(n)=\frac{1+g(n)}2$
So, for example, $g(12)=g(2^2)g(3^1)= \frac{2^5+1}{2+1}\frac{3^3+1}{3} = 11\cdot 7 = 77$ So when $n=12$, your sum is $S(12)=\frac{77+1}{2}=39$.
For $n=143$, $g(143)=g(11)g(13)=\frac{11^3+1}{11+1}\frac{13^3+1}{13+1} = 111\cdot 157 = 17427$ and your sum is $S(143)=\frac{17427+1}{2}=8714$.