Nash equilibria for "presidential election" game 
Suppose, in a country there are $m$ different social issues, positions on which are being indexed with numbers $[-1; 1]$, with radicals on the opposing ends and moderates in the center. In this country there are also $n$ candidates running for president. Before they run, they have to choose their position on all those social issues by placing themselves into some point of this $[-1;1]^m$ «political compass». Suppose they have chosen their positions $a_1, … , a_m$ (as vectors in $\mathbb{R}^n$). Then define $Pref(x)$ as the set of the preferable candidates for voters with position $x$ (the candidates, the Euclidean distance from whose position to $x$ is minimal). Suppose, the voter positions are uniformly distributed on $[-1;1]^m$ and that each voter in the position $x$ chooses one of the candidates from $Pref(x)$ with equal probability. (So the total percentage of votes received by $k$-th candidate is $\int_{[-1;1]^m} \frac{I_{Pref(x)}(k)}{2^m|Pref(x)|}dx$). After the votes are counted, the president is chosen at random with equal probability from those candidates, that received the maximal numbest of votes. The candidates do not have their own opinion on those $m$ issues and only want to win the election (thus their final payoff is the exact probability of them winning). Is there some sort of classification of Nash equilibria in this class of games (for different $m$-s and $n$-s)?

For $n = 1$, every position is a Nash equilibrium as the only candidate will win the election no matter what they do. 
For $n = 2$ both candidates should take the extremely moderate position that is $0$. That is due to the «geometric» fact, that if the positions of two candidates are distinct, then the one who is closer to the center wins. So if $a_1 \neq 0$, then the second candidate can adopt the position $\frac{a_1}{2}$ to win the election for sure, and vice versa.
However, I do not know, how to deal with the general problem.
 A: Several results for $m=1$ case:

Claim 1: For $m=1, n=3$ there is no Nash Equilibrium (N.E.).

Let $x_1 \le x_2 \le x_3$ be the candidates' position on the single $m=1$ issue.  Suppose there is a unique leftmost candidate, i.e. $x_1 < x_2$, then that candidate $1$ can do better by moving just a tiny bit to the right without passing the others (while the others don't move).  So any N.E. must have $x_1 = x_2$.  By considering the rightmost candidate, the same logic says any N.E. must have $x_2 = x_3$.  So the only case still left is $x_1 = x_2 = x_3$, but this isn't a N.E. either: any of them can increase their chance from $\frac13$ to just under $\frac12$ by moving just a bit off-center (while the others don't move).
Remark: For $m> 1, n=3$, I used to think the proof also works because we can just pick any one of the $m$ dimensions (issues) and the above argument applies for just that dimension.  However, on further thought, it's not so simple.  The whole dividing line would "tilt" and it is not clear that a tiny rightward movement by a unique leftmost candidate will always help them.
UPDATE: 

Claim 2: For $m=1, n$ even, there is a N.E.: the candidates come in $n/2$ pairs and the $n/2$ distinct positions are $(\frac1n, \frac3n, \frac5n, \cdots, \frac{n-1}{n})$ - each position occupied by $2$ candidates.

In the given solution, each candidate-pair owns a segment of $\frac1n$ to their left and a segment of $\frac1n$ to their right, so each candidate-pair has prob $\frac2n$ of winning, i.e. each has $\frac1n$ win prob.  Now suppose some candidate moves to point $P$.  If $P \in $ the leftmost (or rightmost) segment, the new win prob is $\le \frac1n$, and if $P \in$ some internal segment, the new win prob $= \frac1n$.  So the given solution is a N.E.

Claim 3: For $m=1, n=4$ the above N.E. is unique.

By the logic used in Claim 1, there cannot be a unique leftmost (nor rightmost) candidate, so when $n=4$ they must come in pairs.  The two pairs' two positions segment the interval into $3$ lengths $a,b,c$ and working through the N.E. math it is easy to show that N.E. can only happen when $2a = b = 2c$, so $(\frac14, \frac14, \frac34, \frac34)$ is the only N.E.
Remark: Still thinking about $n$ odd, and other solutions for $n$ even $\ge 6$, and $m > 1$ etc.
