As others have pointed out, when dealing with congruences the concept of a negative number is meaningless (as is the concept of a positive number). Judging from the comments you are, in addition to this, asking about the solvability of the congruence $m^2\equiv -1\pmod q$, where $q$ is some odd integer. Here the theory says that this congruence has solutions, if and only if all the prime divisors of $q$ are congruent to $1\pmod 4$.
For example, when $q=7$ there are no solutions, because $7\not\equiv1\pmod4$, but when $q=13$, there are solutions $m\equiv\pm5$. When $q$ is a prime, $m=\pm((q-1)/2)!$ are the only non-congruent solutions.
The general theory goes via the usual route: solve it for prime number moduli => solve for prime power moduli (by "lifting") => solve for general moduli with the aid of Chinese remainder theorem. The number of non-congruent solutions depends on the number of prime factors of $q$.
[Edit: In response to Thomas' comment.] If $m^2\equiv-1\pmod q$, then this can be lifted to a solution modulo $q^k$ for all positivie integers $k$. Lifting means that if we have, for some positive integer $k$, a solution $m_k$ such that $m_k\equiv m\pmod q$ and $m_k^2\equiv -1\pmod{q^k}$, then we can find an integer $m_{k+1}$ such that $m_{k+1}=m_k+aq^k$ and $m_{k+1}^2\equiv -1\pmod{q^{k+1}}$. This is not difficult, because we know that $m_k^2=-1+bq^k$ for some integer $b$, and can use this to solve $a$ from the
congruence
$$
m_{k+1}^2=m_k^2+2am_kq^k+a^2q^{2k}\equiv-1+(b+2am_k)q^k\pmod{q^{k+1}},
$$
because this is equivalent to the linear congruence
$$
b+2am_k\equiv0\pmod q.
$$
Here $\gcd(2m_k,q)=1$ so the solutions $a$ of this congruence form a unique residue class modulo $q$.
As an example, let us lift the solution $m=m_1=5$ of the congruence $m^2\equiv -1\pmod{13}$
to a square root of $-1$ modulo $13^2$. Here $m_1^2=25=-1+2\cdot13$, so $b=2$. We want to solve the linear congruence
$$
b+2am_1=2+2\cdot5a\equiv0\pmod{13}.
$$
The usual method for solving a linear congruence yields $a\equiv5\pmod{13}$ as the solution. Therefore $m_2=5+13a=5+5\cdot13=70$ should be a solution. Indeed,
$$
m_2^2=4900=-1+4901=-1+13\cdot377=-1+13^2\cdot29\equiv-1\pmod{13^2}.
$$
[/Edit]
[Edit^2: An example on using the CRT]
Assume that we are given the task to solve the equation
$$
m^2\equiv-1\pmod{2873}.\tag{1}
$$
The process begins by factoring $2873$. If you know this an advance, you are lucky, because otherwise this could be tricky. We simply observe that
$2873=13^2\cdot17$. It is our lucky day, because all the prime factors are $\equiv 1\pmod 4$ (of course I reverse engineered this example a bit). The idea is that we next solve the
congruences
$$
m^2\equiv-1\pmod{13^2}\tag{2}
$$
and
$$
m^2\equiv -1\pmod{17}\tag{3}$$
separately. In the previous example I showed how to lift the "guessable" solution $5^2\equiv1\pmod{13}$ to a solution $m=70$ of equation $(2)$, so let's reuse that.
We observe that $-70\equiv99\pmod{13^2}$ is then the other solution of $(2)$.
Finding the solutions of $(3)$ is also easy, because $17$ is a relatively small number.
There are several methods. Either we simply observe that $m=\pm4$ are solutions. Or, as above, we can calculate that $(17-1)/2=8$ and $8!=40320\equiv 13\equiv-4\pmod{17}$. We can also use the quicker (for primes a bit larger than $17$) but non-deterministic method that for any integer $a, 0<a<17$, the power $x=a^{(p-1)/4}=a^4$ is a solution of either the equation $x^2\equiv1$ or $x^2\equiv-1$. The non-deterministic part is that we don't know which it is. Let's try. With $a=2$ we get $x=16\equiv-1$, which is not a solution of $(3)$.
But with $a=3$ we get $x=81\equiv13\equiv-4$, which we already knew to be a solution. Anyway, we now know that $(3)$ holds, if and only if $m\equiv\pm4\pmod{17}.$
Because $17$ and $13^2$ are coprime (no common prime factors), the Chinese remainder theorem tells us that $m$ satisfies congruence $(1)$, if and only if it satisfies both congruences $(2)$ and $(3)$. We can also conclude from CRT that there is a single residue class modulo $2873$ that satisfies e.g. both congruences $m\equiv70\pmod{13^2}$ and $m\equiv 4\pmod{17}$. The extended Euclidean algorithm gives us a method for finding integers $u$ and $v$ such that $$u\cdot 169+v\cdot 17=1,$$ for example $u=-1,v=10$. What this means is that the number $e_1=v\cdot17=170$ is congruent to $1$ modulo $13^2$ to $0$ modulo $17$, and OTOH the number $e_2=u\cdot169=-169$ is congruent to $0$ modulo $13^2$ and to $1$ modulo $17$.
Therfore
$$m_1=70\cdot e_1+4\cdot e_2=70\cdot170-4\cdot169=11224\equiv2605\pmod{2873}$$ is congruent to $70$ modulo $13^2$ and to $4$ modulo 17. So it should be a solution to $(1)$. With the aid of a calculator we can check
$$
2605^2=6786025\equiv-1\pmod{2873},
$$ so this is, indeed a solution. Using other solution to congruences $(2)$ and $(3)$ we get a total of of four non-congruent solutions:
$$
m_2=99\cdot e_1+4\cdot e_2=16154\equiv1789\pmod{2873}$$
is congruent to $99$ modulo $13^2$ and to $4$ modulo $17$. Using the residue class $-4$ modulo $17$ gives us two more solutions
$$
m_3=70\cdot e_1-4\cdot e_2=12576\equiv 1084\pmod{2873},$$
and
$$m_4=99\cdot e_2-4\cdot e_2=17506\equiv 268\pmod{2873}.$$
In modular arithmetic (or more generally, in a group), $a^{-1}$ does not mean $\frac{1}{a}$ in the usual sense of $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, etc. (i.e. $67^{-1}$ does not mean $1 \div 67 \approx 0.0149$). In modular arithmetic, $a^{-1}$ means the multiplicative inverse of $a$. That is, it is the unique element in $\{1, \dots, 118\}$ such that $aa^{-1} = a^{-1}a \equiv 1 \pmod{119}$; note, the existence of an inverse for any non-zero element relies on the fact that the modulus is prime. If the modulus is not prime, the only non-zero elements which have multiplicative inverses are those which are coprime to the modulus.
Note: As Bill Dubuque outlines in his answer, provided $a$ is coprime to the modulus, you can treat $a^{-1}$ as $\frac{1}{a}$, but it represents a different value than it does in say $\mathbb{R}$.
Let $x$ be the multiplicative inverse of $67$ modulo $119$ (i.e. $x \equiv 67^{-1} \pmod{119}$). Then you are looking to solve $67x \equiv 1 \pmod{119}$. By definition, $67x \equiv 1 \pmod{119}$ means that $119 \mid (67x - 1)$ so there is some integer $y$ such that $67x - 1 = 119y$, or written differently $67x + 119y = 1$. Note, if $67$ and $119$ weren't coprime, this equation would have no integer solutions. This is a diophantine equation that you can solve by using the Euclidean algorithm and back substitution. Have you seen this before?
Best Answer
You can represent equivalence classes any way you want to. If you're asking whether you can give ${\mathbb Z}/p{\mathbb Z}$ the structure of an ordered field, the answer is no. To see this, consider that $1$ is positive, so any sum of $1$'s is positive, so everything is positive.