JavaScript (ES6), 190 bytes

(m,n)=>m.map((a,r)=>a.map((_,c)=>f(r,c,[0],0)),o=f=(x,y,s,t)=>s[n]?o>t?0:o=t:s.indexOf(w=x+","+y)<0&&m[y]&&(v=m[y][x])<1/0&&f(x+1,y,s=[...s,w],t+=v)+f(x,y+1,s,t)+f(x-1,y,s,t)+f(x,y-1,s,t))|o 

Explanation

Takes the matrix as an array of arrays.

Starts from each square then uses a recursive function to test every possible combination. This is a brute-force approach, but it finishes almost instantly for the first test case on my machine.

(m,n)=> m.map((a,r)=> // for each row a.map((_,c)=> // for each column f(r,c,[0],0) // start checking paths from the coordinate of the square ), o= // o = output number (max total) f=(x,y,s,t)=> // recursive function f, x & y = current square, t = total // s = array of used squares (starts as [0] so length = +1) s[n]? // if we have used n squares o>t?0:o=t // set o to max of o and t :s.indexOf(w=x+","+y)<0&& // if the square has not been used yet m[y]&&(v=m[y][x])<1/0&& // and the square is not out of bounds // ( if value of square is less than Infinity ) // Check each adjacent square f(x+1,y, s=[...s,w], // clone and add this square to s t+=v // add the value of this square to the total ) +f(x,y+1,s,t) +f(x-1,y,s,t) +f(x,y-1,s,t) ) |o // return output 

Test

var solution = (m,n)=>m.map((a,r)=>a.map((_,c)=>f(r,c,[0],0)),o=f=(x,y,s,t)=>s[n]?o>t?0:o=t:s.indexOf(w=x+","+y)<0&&m[y]&&(v=m[y][x])<1/0&&f(x+1,y,s=[...s,w],t+=v)+f(x,y+1,s,t)+f(x-1,y,s,t)+f(x,y-1,s,t))|o

<textarea rows="7" cols="40" id="Matrix">10 -7 11 7 3 31 33 31 2 5 121 15 22 -8 12 10 -19 43 12 -4 54 77 -7 -21 2 8 6 -70 109 1 140 3 -98 6 13 20</textarea><br /> N = <input type="number" id="N" value="6" /><br /> <button onclick="result.textContent=solution(Matrix.value.split('\n').map(x=>x.split(' ').map(z=>+z)),N.value)">Go</button> <pre id="result"></pre>