We have $n=2x+1$ and $n=3y+2$. Hence, $$2x+1=3y+2 \Rightarrow 2x-3y=1.$$We know $-3y\equiv 2y \pmod 5$, so $2x+2y \equiv 1 \pmod 5$. Checking all the cases, we conclude $\boxed{x+y \equiv 3 \pmod 5}$.

Quote: Checking all the cases, we conclude $\boxed{x+y \equiv 3 \pmod 5}$. Alternatively, multiply by $2^{-1}$ on both sides.

AshAuktober wrote: Quote: Checking all the cases, we conclude $\boxed{x+y \equiv 3 \pmod 5}$. Alternatively, multiply by $2^{-1}$ on both sides. Yeah, that's essentially what I did (use the fact that $5$ is odd), but remember this is an olympiad for 7th grade

This stuff is probably more suited for HSM based on the difficulty. Also does this comp require proofs or is it short answer? If the latter, then one could simply notice that $\tfrac{5}{2}=2R1$ and $\tfrac{5}{3}=1R2,$(where $\tfrac{a}{b}=cRd$ if $a=bc+d$ and $d<b$, so $c$ is the quotient and $d$ is the remainder) so the answer is $1+2=\boxed{3}.$

Let a be a number such that a=2x+1=3y+2. We have that x+y=(5a-7) : 6, so 6(x+y)=5a-7, where we get that x+y is 3 (mod 5).