Solution:
Answer: 117690
We will find the number of such triples with $a<b<c$. The answer to the original problem will then be six times what we will get. By symmetry, the number of triples $(a, b, c)$ with $a+b+c>0$ is equal to the number of those with $a+b+c<0$. Our main step is thus to find the number of triples with sum $0$.

If $b=0$, then $a=-c$, and there are $31$ such triples. We will count the number of such triples with $b>0$ since the number of those with $b<0$ will be equal by symmetry.

For all positive $n$ such that $1 \leq n \leq 15$, if $a=-2n$, there are $n-1$ pairs $(b, c)$ such that $a+b+c=0$ and $b>0$, and for all positive $n$ such that $1 \leq n \leq 16$, if $a=-2n+1$, there are also $n-1$ such pairs $(b, c)$. In total, we have $1+1+2+2+3+3+\ldots+14+14+15=225$ triples in the case $b>0$ (and hence likewise for $b<0$.)

In total, there are $31+225+225=481$ triples such that $a<b<c$ and $a+b+c=0$. Since there are $\binom{63}{3}=39711$ triples $(a, b, c)$ such that $-31 \leq a<b<c \leq 31$, the number of triples with the additional restriction that $a+b+c>0$ is $\frac{39711-481}{2}=19615$. So the answer to the original problem is $19615 \times 6=117690$.