The biggest degree of a map between fixed surfaces Let $S_1$ and $S_2$ be compact surfaces (real manifolds of dimension 2). None of them is a sphere.
Question: What is the biggest degree of a smooth map from $S_1$ to $S_2$?
Comment 1. I have a conjecture. Integer part of
 $\frac{ \chi (S_1) } { \chi(S_2) }$.
Denote this number as k. At least I can construct a map of degree $k$. Each surface (which is not a sphere) has an $n$-fold covering for any $n$. So we can consider $\tilde{S}_2 -$ $k$-fold covering of $S_2$. It is easy to see that $S_1$ has not less handles than $\tilde{S}_2$ does. So there is a map of degree 1 from $S_1$ to $\tilde{S}_2$ which contracts some handles.
Comment 2. If we require $S_1$ and $S_2$ to be Riemann surfaces and would consider holomorphic maps, than this result is trivial from Riemann-Hurwitz formula.
 A: Jason DeVito's answer dealt with the case where $S_1 = S^1\times S^1$ and Jesus RS's answer dealt with the case where $S_2 = S^1\times S^1$. You said that neither surface is $S^2$, so this answer addresses the remaining case: $S_1 = \Sigma_g$ and $S_2 = \Sigma_h$ with $g, h \geq 2$.

The Gromov norm $\|\cdot\|$, also known as simplicial volume, is a useful tool for tackling questions about degree due to the following fact:

Let $M$ and $N$ be closed, oriented manifolds of the same dimension and $f : M \to N$ a continuous map. Then $|\deg f|\|N\| \leq \|M\|$.

The Gromov norm of $\Sigma_g$, for $g \geq 2$, is $\|\Sigma_g\| = 4g - 4$ (see this answer). So for any $f : \Sigma_g \to \Sigma_h$ with $g, h \geq 2$, we have $|\deg f|(4h - 4) \leq (4g - 4)$ and therefore
$$|\deg f| \leq \left\lfloor\frac{g-1}{h-1}\right\rfloor  = \left\lfloor\frac{2 - 2g}{2 - 2h}\right\rfloor = \left\lfloor\frac{\chi(\Sigma_g)}{\chi(\Sigma_h)}\right\rfloor =: L.$$
In particular, if $g < h$, then $f$ has degree zero which agrees with Mike Miller's answer.
This upper bound is actually achieved, i.e. there is a map $f : \Sigma_g \to \Sigma_h$ of degree $L$. This can be seen by combining the following two facts:

  
*
  
*If $a \geq b$, then there is a degree one map $\Sigma_a \to \Sigma_b$ (write $\Sigma_a = \Sigma_b\#\Sigma_{a-b}$ then crush $\Sigma_{a-b}$ to a point).
  
*There is a $k$-sheeted covering map $\Sigma_a \to \Sigma_b$ if and only if $\chi(\Sigma_a) = k\chi(\Sigma_b)$, i.e. $a = k(b - 1) + 1$ (see Example $1.41$ of Hatcher's Algebraic Topology).
  

As $L(h-1) + 1 \leq g$, there is a degree one map $\Sigma_g \to \Sigma_{L(h-1)+1}$ by the first point. By the second point, there is an $L$-sheeted covering map $\Sigma_{L(h-1)+1} \to \Sigma_h$. The composition of these two gives a map $\Sigma_g \to \Sigma_h$ of degree $L$.
In conclusion, you are correct.
A: At least for mappings into a torus we can produce all integers as degrees. Let $T=S^1\times S^1$ be the torus. We see $S^1\subset\mathbb C$ and then we have
$f_k:T\to T$ given by $f_k(z,w)=(z^k,w)$, which has degree $k$, for any $k\in \mathbb Z$. On the other hand, for any orientable surface $S_g$ with $g>1$ holes, one can collapse all the surface but a nbhd of the first hole onto a point, which gives a torus $T$, so getting a degree $1$ map $h:S_g\to T$. Now compose to get a map $h_k=f_k\circ h:S_g\to T$ of degree $k$. The latter is not smooth, but a good enough approximation will have the same degree. 
On the other hand, suppose $h:T\to S$ is a map of degree $d\ne0$ into some other surface $S$. Then the composite $h\circ f_k:T\to S$ has degree $kd$. We see that the bound exists only if all mappings $h:T\to S$ have degree $0$ (and the bound is $0$). 
A: Let me confirm your conjecture when $S_1 = T$ is a torus.  Then, the conjecture asserts that every map $f:T\rightarrow S$ is degree $0$ if $S$ is not a torus or a sphere.
(As Jesus RS pointed out, if $S$ is a torus, you can get maps of any degree.  It is also true that if $S = S^2$, then you get can get any degree.)
This MSE question proves that for every map $f:T\rightarrow S$, $f_\ast:H_1(T)\rightarrow H_1(S)$ is not injective.  It follows that the same is true if we change coefficient from $\mathbb{Z}$ to $\mathbb{Q}$.
Dualizing, we see that no $f$ induces a surjection from $f^\ast:H^1(S;\mathbb{Q})\rightarrow H^1(T;\mathbb{Q} = \mathbb{Q}^2$.  But, since $f^\ast$ is linear and not surjective, the image must lie on a line in $\mathbb{Q}^2$.
Since the cup product is graded anti-commutative, it follows that for any $v,w\in H^1(S;\mathbb{Q})$, that $f^\ast(v)\cup f^\ast(w) = 0 \in H^2(T)$.
Now, picking $v$ and $w$ so that $v\cup w$ generates $H^2(S;\mathbb{Q})$, we see that $f^\ast:H^2(S;\mathbb{Q})\rightarrow H^2(T;\mathbb{Q})$ must be the $0$ map on $H^2$.  Thus, $f$ has degree $0$.
A: If $g < h$, maps $M_g \to M_h$ necessarily have degree zero.
Suppose we had a map $f: M_g \to M_h$ with nonzero degree $d$. Let $K = \text{Im}(f_*: \pi_1(M_g) \to \pi_1(M_h))$, and lift $f$ to $\tilde f: M_g \to M_K$, the surface corresponding to the covering $p: M_K \to M_h$ with $p_*(\pi_1(M_K)) = K$. $M_K$ must be compact, lest the map on top homology factor through $H_2(M_K) = 0$ and force $d=0$; and so we have a surjection $\pi_1(M_g) \to \pi_1(M_K)$. By an Euler characteristic argument the genus of $M_K$ is greater than the genus of $M_h$; call it $g'$. $\pi_1(M_g)$ is generated by $2g$ elements, so such a surjection would generate $H_1(M_K) = \Bbb Z^{2g'}$ with $2g$ elements. But this is impossible. So $f$ must have had degree zero all along.
Jason DeVito sent me an alternative cohomological argument: a map $f: M_g \to M_h$ induces a map $f^*: H^1(M_h) \to H^1(M_g)$; because of the ranks involved, this has some $x$ in the kernel. Poincare duality (or knowing the cohomology ring) gives you some $y$ such that $x \smile y$ is nonzero in $H^2(M_h)$; then $f^*(x \smile y) = f^*(x) \smile f^*(y) = 0$, so the map must have degree zero.
