Basically the law of equivalence wants you to balance the equivalents that are involved in your reaction. I'd like you to view you reactions in two parts, the oxidation and reduction halves. I see:
$$\begin{align} \ce{I- -> I2} \tag{oxidation} \\
\ce{IO3- -> I2}\tag{reduction} \end{align}$$
In other words, we've just discovered a (reverse disproportionation?) comproportionation reaction. The change in the oxidation state of iodine in the oxidation part is $-1 \to 0 $ ($\Delta_{o.s.} = 1$) and for the reduction half it's $+5 \to 0$ ($\Delta_{o.s.} = 5$). So far so good.
Usually, the questioner would provide you with the number of moles of each species and you're done with number of moles times the n-factor. Nevertheless, let's try to give it a shot. I'm balancing this equation:
$$\ce{1KIO3 + 5KI + 3H2SO4 \to 3I2 + 3K2SO4 + 3H2O}$$
This means for $1$ mole of $\ce{KIO3}$ to react, you'll need $5$ moles of $\ce{KI}$. As this is a proportion, converting it to an absolute value, I'd say you've $1n$ moles of $\ce{KIO3}$ and $5n$ moles of $\ce{KI}$.
So if I try to equate the gram-equivalents, I'll have $1n \times 5 = 5n \times 1$. In other words, $\text{Eq.}_{\ce{KIO3}} = \text{Eq.}_{\ce{KI}}$
I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products.
Law of equivalence is even more fun, for a balanced chemical reaction:
$$ \ce{aA + bB -> cC + dD}$$
Equivalents of every species will be equal. Mathematically,
Equivalents of A = Equivalents of B = Equivalents of C = Equivalents of D
will hold.
Why aren't the gram equivalents of $\ce{H2SO4}$ equal to that of water?
You need to strip your equation of the spectating species. For instance $\ce{H2SO4}$ is primarily here because it acts as a medium-provider, it makes the solution acidic. Neither $\ce{SO4^2-}$ nor $\ce{H+}$ underwent any oxidation/reduction so applying the equivalent concept makes no sense in this case.