1. We need the remainder of $19^{200}+23^{200}$ when divided by $49$.

So we work modulo $49$.

2. Reduce the bases modulo $49$: $19 \equiv 19 \pmod{49}, \qquad 23 \equiv 23 \pmod{49}.$ A better observation is: $19= -30, \qquad 23=-26 \pmod{49}$ but an even simpler route is to use powers.

3. Compute smaller powers modulo $49$.

For $19$: $19^2=361 \equiv 361-343=18 \pmod{49}.$ Then $19^4 \equiv 18^2=324 \equiv 324-294=30 \pmod{49},$ $19^8 \equiv 30^2=900 \equiv 900-882=18 \pmod{49}.$ So powers are cycling. In fact, $19^2 \equiv 18 \pmod{49}, \qquad 19^6=19^2\cdot 19^4 \equiv 18\cdot 30=540 \equiv 1 \pmod{49}.$ Thus, $19^6 \equiv 1 \pmod{49}.$ Hence $19^{200}=19^{6\cdot 33+2}=(19^6)^{33}\cdot 19^2 \equiv 1^{33}\cdot 18=18 \pmod{49}.$

4. For $23$: $23^2=529 \equiv 529-490=39 \pmod{49}.$ Then $23^4 \equiv 39^2=1521 \equiv 1521-1470=51 \equiv 2 \pmod{49}.$ Also, $23^6=23^2\cdot 23^4 \equiv 39\cdot 2=78 \equiv 29 \pmod{49},$ not yet $1$. So let us continue more carefully.

A better approach: since $23^2 \equiv 39 \pmod{49},$ $23^3 \equiv 23\cdot 39=897 \equiv 897-882=15 \pmod{49},$ $23^6 \equiv 15^2=225 \equiv 225-196=29 \pmod{49},$ $23^{12} \equiv 29^2=841 \equiv 841-833=8 \pmod{49}.$ This is not simplifying quickly.

So instead use the fact that $\varphi(49)=42$, and since $\gcd(23,49)=1$, $23^{42}\equiv 1\pmod{49}.$ Now $200=42\cdot 4+32,$ so $23^{200}\equiv 23^{32}\pmod{49}.$ Let us compute by repeated squaring: $23^2\equiv 39,$ $23^4\equiv 39^2=1521\equiv 2,$ $23^8\equiv 2^2=4,$ $23^{16}\equiv 4^2=16,$ $23^{32}\equiv 16^2=256\equiv 256-245=11 \pmod{49}.$ Therefore, $23^{200}\equiv 11 \pmod{49}.$

5. Add the two remainders: $19^{200}+23^{200} \equiv 18+11=29 \pmod{49}.$

Therefore, the remainder is $\boxed{29}.$

6. Comparison with stored answer: Stored correct answer = $29$. Our derived answer also = $29$. So they agree.