$A$ covering space of $B$ of degree $n$ iff $\chi(A) = n\chi(B)$. If $A$, $B$ are two orientable surfaces, do we have that $A$ is a covering space of $B$ of degree $n$ if and only if$$\chi(A) = n\chi(B)?$$If this is the case, then does it follow there exist degree $n$ covers by the torus of itself of arbitrary degree?
 A: I'm going to solve this for the case of compact surfaces, as nothing in the problem suggests that OP is asking for anything else. Let $T_g$ denote the $g$-holed torus. Its Euler characteristic is $2-2g$. For any $g$ this is an integer multiple of $2$, which is the Euler characteristic of the sphere; but that does not make the $g$-holed torus a $1-g$-fold cover of the sphere. Indeed, since $1-g$ is nonpositive, such a cover doesn't even make sense. 
So I'm going to assume that the OP meant "compact surfaces" and $n > 0$. 
This breaks into two cases:


*

*$T_1$, with $\chi = 0$. In this case, for any $n>0$, we have $n \chi = 0$, so we're asking if $T_1$ is an $n$-fold cover of $T_1$. Using the usual representation of $T_1$ as $S^1 \times S^1$, we have the map $(u, v) \mapsto ((n\cdot u) \bmod 2\pi, v)$, which is indeed an $n$-fold cover of the torus by itself. So in this case the answer is "yes." 

*$T_g$, where $g > 1$. 
In this case, if there were a covering of the base $T_g$, with $\chi_g = 2-2g$, the cover would have to be $T_k$ with $\chi_k = (2-2g)n = 2 - 2k$. Solving, we find $2n - 2gn = 2 - 2k$, so $k = (gn+1-n)$. Is there always such a cover?  Does $T_k$, with $k = gn+1-n$, always form an $n$-fold cover of $T_k$? (At least for $n > 0$.) 
Yes. For $n = 1$, this is obvious: it covers itself. For $n > 1$, here's a construction: 
Start with a standard 3-holed torus, lying on the plane like three bagels that got stuck together during baking, with the holes looking like this from above:
 OOO

Cut it in half down the middle, so you have a "torus with horns" that looks like this:
 OC

(I'm hoping that this sort of schematic picture makes sense to you!)
Let's call that thing $H$ (for "horns"). If you take two copies of $H$, you can clearly rotate one by 180 degrees and glue them together to reconstruct $T_3$. If you take 3 copies of $H$, slightly shorten the horns on each one, and arrange them at 120-degrees from each other, you get a 4-holed torus. If you take 4 copies of $H$, with even shorter horns, and arrange them at 90 degrees from each other, you get a 5-holed torus that schematically looks like this:
   O
  OOO
   O

with the middle "O" being made up from the four sets of horns. 
OK, back to the question at hand: take a g-holed torus, and arrange it like our 3-holed torus, and chop off half of the rightmost hole, so you get
 OOO ... OC

Arrange $n$ copies of this in a "star" pattern, with angle $\frac{2\pi}{n}$ between each point of the star, and with the "horns" in the middle forming a circle. The result is a surface that forms an $n$-fold cover of the original surface, the covering map being essentially $z \mapsto z^n$, where $z$ is a complex coordinate in the plane of the drawing. 
That surface has $n$ copies of the $g-1$ holes of our chopped-off $g$-holed torus, i.e., has $n(g-1)$ holes in its "arms", and one more hole in the center, for a total of $ng - n + 1$ holes. So we see that we've produced an $n$-fold covering of the $g$-holed torus. 
If all this is too horribly cut-and-paste and geometric for you, here's an alternative: represent $T_g$ as an n-fold connect sum $T_g = T_1 \# T_1 \# \ldots \# T_1$. $\pi_1$ of the first $T_1$ is generated by a meridian $a$ and a longitude $b$, and under inclusion, each of these represents an element of $\pi_1 T_g$. Consider the subgroup $K$ of $\pi_1 T_g$ consisting of words in which $a$ appears only with powers that are a multiple of $n$, and construct the regular cover corresponding to $K$. This will have degree $n$. The covering space will be a compact orientable surface, and will have the right Euler characteristic, and hence will have to be $T_k$ for $k = gn + 1 - n$. 
