Just opposite!

If $n$ is odd, then every vertex of the complete graph $K_n$ has even valency, and hence an Eulerian cycle is possible. Clearly, using the complete graph will give the maximum edge sum. Thus every vertex contributes to the value of $n-1$ edges, and hence $f(n) \; = \; (n-1)\sum_{i=1}^n i \; = \; \tfrac12n(n+1)(n-1)$ Thus $f(n) \equiv 0 \pmod{n}$ , so if $f(n) \equiv 2013 \pmod{n}$ then $2013 \equiv 0 \pmod{n}$ , and hence $n$ divides $2013$ . Since $2013 = 3 \times 11 \times 61$ , there are $2^3 = 8$ divisors of $2013$ , and all of these are odd.

If $n=2$ , then $K_2$ possesses an Euler path, and $f(2) = 3 \equiv 2013 \pmod{2}$ .

If $n\ge4$ is even, then all the vertices of the complete graph $K_n$ have odd valency. To be able to have an Euler path, we need to remove edges from $K_n$ so that $n-2$ vertices have their valencies made even, and we wish to do this in a way which minimizes the reduction to the edge sum. We do this by removing one edge from each of the vertices $1,2,...,n-2$ ; since $n$ is even, this can be done without any repetition of vertices, for example by removing $(1,2)$ , $(3,4)$ , ..., $(n-3,n-2)$ . This means that $f(n) \; = \; \tfrac12n(n+1)(n-1) - \tfrac12(n-2)(n-1) \; = \; \tfrac12(n-1)\big[n(n+1) - (n-2)\big] \; = \; \tfrac12(n-1)(n^2 + 2) \; = \; \tfrac12n^2(n-1) + n - 1$ and so we note that $f(n) \equiv -1\pmod{n}$ this time. If $f(n) \equiv 2013 \pmod{n}$ then $2013 \equiv -1 \pmod{n}$ , so that $n$ divides $2014 = 2 \times 19 \times 53$ . Thus there are $4$ even divisors of $2014$ , and $3$ of these are greater than $2$ .

Thus there are $8 + 1 + 3 = \boxed{12}$ positive integers $n$ such that $f(n) \equiv 2013 \pmod{n}$ (the eight divisors of $2013$ , and the four even divisors of $2014$ ).