jmc

Let's begin by ignoring the condition that $x<y$. Instead, suppose $x,y$ are any two (not necessarily distinct) numbers between $1$ and $100$, inclusive. We want $i^x+i^y$ to be real.

Any pair of even numbers will work, as both $i^x$ and $i^y$ will be real; there are $50\cdot 50=2500$ such pairs. Note that among these pairs, exactly $50$ of them satisfy $x=y$.

We have two other possibilities; (a) $i^x=i$ and $i^y=-i$, or (b) $i^x=-i$ and $i^y=i$. Note that there are $25$ numbers $n$ for which $i^n=i$ (namely, $n=1,4,\dots ,97$), and there are $25$ numbers $n$ for which $i^n=-i$ (namely $n=3,7,\dots ,99$). Therefore, there are $25\cdot 25=625$ desirable pairs in case (a), and similarly, there are $625$ desirable pairs in case (b), resulting in an additional $625+625=1250$ pairs. Note that none of these pairs satisfy $x=y$.

Therefore, there are a total of $2500+1250=3750$ pairs $(x,y)$ with $1\le x,y\le 100$ such that $i^x+i^y$ is a real number. Now, let's try to determine how many of these satisfy $x<y$. First of all, let's remove the $50$ pairs with $x=y$, leaving us with $3700$ pairs. Among these $3700$ pairs, we know that exactly half of them satisfy $x<y$ and the other half satisfy $x>y$ by symmetry. Therefore, the answer is $3700/2=\boxed{1850}$.